Question

identify the graph of $f(x) = \frac{10 - 10e^x}{x^2}$?

Explanation:

Step1: Analyze the function's domain

The function \( f(x)=\frac{10 - 10e^{x}}{x^{2}} \) has a denominator \( x^{2} \), so \( x eq0 \). The graph should have a vertical asymptote at \( x = 0 \).

Step2: Analyze the function's behavior as \( x\to0^{+} \)

As \( x\to0^{+} \), \( e^{x}\to1 \), so numerator \( 10 - 10e^{x}\to0^{-} \), and \( x^{2}\to0^{+} \). So \( f(x)=\frac{\text{negative small}}{\text{positive small}}\to-\infty \).

Step3: Analyze the function's behavior as \( x\to0^{-} \)

As \( x\to0^{-} \), \( e^{x}\to1 \), numerator \( 10 - 10e^{x}\to0^{-} \), and \( x^{2}\to0^{+} \) (since square of negative is positive). So \( f(x)=\frac{\text{negative small}}{\text{positive small}}\to-\infty \).

Step4: Analyze the function's behavior as \( x\to\pm\infty \)

As \( x\to\pm\infty \), \( e^{x}\to\infty \) (for \( x\to\infty \)) or \( e^{x}\to0 \) (for \( x\to-\infty \)). For \( x\to\infty \): numerator \( 10 - 10e^{x}\to-\infty \), denominator \( x^{2}\to\infty \), so \( f(x)\to-\infty \). For \( x\to-\infty \): numerator \( 10 - 10e^{x}\to10 - 0 = 10 \), denominator \( x^{2}\to\infty \), so \( f(x)\to0^{+} \).

Now, looking at the given graphs, the fourth graph (the one with the coordinate system) has behavior: near \( x = 0 \), both sides go to \( -\infty \), as \( x\to\infty \) goes to \( -\infty \), as \( x\to-\infty \) goes to \( 0^{+} \)? Wait, no, maybe I made a mistake. Wait the fourth graph shown: the left and right parts (for \( x < 0 \) and \( x > 0 \)): for \( x > 0 \), as \( x\) approaches 0 from right, the graph goes down (to \( -\infty \)), and as \( x\to\infty \), goes up? Wait no, the red arrows: for \( x > 0 \), the graph is a U - like but wait no, the arrows: on the right side ( \( x>0 \) ), the arrow is going up, on the left side ( \( x < 0 \) ), arrow going down? Wait maybe I misanalyzed. Wait the function is \( f(x)=\frac{10 - 10e^{x}}{x^{2}}=10\frac{1 - e^{x}}{x^{2}} \). Let's re - evaluate:

Wait \( f(x)=\frac{10(1 - e^{x})}{x^{2}} \). For \( x = 1 \): \( f(1)=\frac{10(1 - e)}{1}\approx10(1 - 2.718)=10(-1.718)= - 17.18 \). For \( x=-1 \): \( f(-1)=\frac{10(1 - e^{-1})}{1}\approx10(1 - 0.368)=10(0.632)=6.32 \).

So at \( x = - 1 \), \( f(-1)>0 \); at \( x = 1 \), \( f(1)<0 \). Now, looking at the fourth graph: for \( x < 0 \) (left of y - axis), the graph is below or above? The fourth graph: the left part ( \( x < 0 \) ) has arrow going down (to \( - 2 \) direction) but wait no, the coordinate system: x from - 2 to 2, y from - 2 to 2. Wait maybe the fourth graph is not the right one. Wait the first three graphs are blank except for some symbols, the fourth has the coordinate. Wait maybe the correct graph is the one where for \( x>0 \), as \( x\to0^{+} \), \( f(x)\to-\infty \), \( x\to\infty \), \( f(x)\to-\infty \); for \( x < 0 \), \( x\to0^{-} \), \( f(x)\to-\infty \), \( x\to-\infty \), \( f(x)\to0^{+} \). The fourth graph (the one with the x - y axes) shows: for \( x>0 \), the graph is a U - shape? No, the red arrows: on the right side ( \( x>0 \) ), the graph goes up (arrow up), on the left side ( \( x < 0 \) ), arrow down. Wait maybe I messed up the function. Wait the function is \( f(x)=\frac{10 - 10e^{x}}{x^{2}}=10\frac{1 - e^{x}}{x^{2}} \). Let's take derivative to check monotonicity, but maybe the key is the vertical asymptote at \( x = 0 \), and the sign.

Wait when \( x>0 \): \( e^{x}>1 \), so \( 1 - e^{x}<0 \), so \( f(x)<0 \). When \( x < 0 \): \( e^{x}<1 \), so \( 1 - e^{x}>0 \), so \( f(x)>0 \). Ah! That's a key point. So for \( x>0 \), \( f(x)<0 \); for \( x < 0 \), \( f(x)>0 \). Now, looking at the fourth graph: the right part ( \( x>0 \) ) is below the x - axis (since \( f(x)<0 \)), and the left part ( \( x < 0 \) ) is above the x - axis (since \( f(x)>0 \)). Now, the fourth graph: the left part ( \( x < 0 \) ) has the arrow going down (but if \( f(x)>0 \) for \( x < 0 \), it should be above x - axis), and the right part ( \( x>0 \) ) is above? Wait no, the fourth graph's right part is above y = 0? Wait the fourth graph: the red curves: for \( x>0 \), the curve is above the x - axis (y>0), but we just found \( f(x)<0 \) for \( x>0 \). So maybe I made a mistake in sign. Wait \( f(x)=\frac{10 - 10e^{x}}{x^{2}}=10\frac{1 - e^{x}}{x^{2}} \). For \( x>0 \), \( e^{x}>1 \), so \( 1 - e^{x}<0 \), so \( f(x)<0 \) (correct). For \( x < 0 \), \( e^{x}<1 \), so \( 1 - e^{x}>0 \), so \( f(x)>0 \) (correct). So the graph should have: for \( x < 0 \), \( f(x)>0 \) (above x - axis), as \( x\to0^{-} \), \( f(x)\to-\infty \)? Wait no, earlier analysis: as \( x\to0^{-} \), numerator \( 10 - 10e^{x}\to0^{-} \) (since \( e^{x}\to1 \), so \( 10 - 10e^{x}\to0 \) from below), denominator \( x^{2}\to0^{+} \), so \( f(x)=\frac{\text{negative small}}{\text{positive small}}\to-\infty \). Wait that contradicts the sign from \( 1 - e^{x} \). Wait \( 1 - e^{x} \) when \( x\to0^{-} \): \( x\) is negative, \( e^{x}=e^{-|x|}=\frac{1}{e^{|x|}} \), so \( 1 - e^{x}=1 - \frac{1}{e^{|x|}} \). As \( |x|\to0 \), \( e^{|x|}\to1 \), so \( 1 - \frac{1}{e^{|x|}}\to0^{-} \) (since \( \frac{1}{e^{|x|}}>1 \) when \( |x| < 0 \)? No, \( |x| \) is non - negative, when \( x\to0^{-} \), \( |x|\to0^{+} \), \( e^{|x|}\to1^{+} \), so \( \frac{1}{e^{|x|}}\to1^{-} \), so \( 1 - \frac{1}{e^{|x|}}\to0^{+} \). Oh! I made a mistake earlier. \( e^{x} \) when \( x\to0^{-} \): \( x=-a \), \( a\to0^{+} \), \( e^{-a}=\frac{1}{e^{a}}\to1^{-} \), so \( 10 - 10e^{x}=10(1 - e^{-a})\to10(1 - 1^{+})=10(0^{-}) \)? No, \( 1 - e^{-a}=1-\frac{1}{e^{a}}=\frac{e^{a}-1}{e^{a}} \), as \( a\to0^{+} \), \( e^{a}-1\to0^{+} \), \( e^{a}\to1^{+} \), so \( 1 - e^{-a}\to0^{+} \). So numerator \( 10 - 10e^{x}=10(1 - e^{x})\to0^{+} \) when \( x\to0^{-} \). And denominator \( x^{2}\to0^{+} \). So \( f(x)=\frac{\text{positive small}}{\text{positive small}}\to+\infty \) when \( x\to0^{-} \). Ah! I had a sign error earlier. Let's re - do the limit:

For \( x\to0^{-} \):

\( e^{x}=e^{-|x|}=\frac{1}{e^{|x|}} \), so \( 1 - e^{x}=1-\frac{1}{e^{|x|}}=\frac{e^{|x|}-1}{e^{|x|}} \). As \( |x|\to0^{+} \), \( e^{|x|}-1\sim|x| \) (Taylor series), so \( 1 - e^{x}\sim\frac{|x|}{1}=|x|>0 \) (since \( |x|>0 \) when approaching 0 from left). So numerator \( 10(1 - e^{x})\sim10|x|>0 \), denominator \( x^{2}=|x|^{2}>0 \). So \( f(x)=\frac{10|x|}{|x|^{2}}=\frac{10}{|x|}\to+\infty \) as \( x\to0^{-} \) (since \( |x|\to0^{+} \)).

For \( x\to0^{+} \):

\( e^{x}\to1^{+} \), so \( 1 - e^{x}\to0^{-} \), numerator \( 10(1 - e^{x})\to0^{-} \), denominator \( x^{2}\to0^{+} \), so \( f(x)=\frac{10(1 - e^{x})}{x^{2}}\to-\infty \) (since negative over positive).

For \( x\to\infty \):

\( e^{x}\to\infty \), so \( 1 - e^{x}\to-\infty \), denominator \( x^{2}\to\infty \), so \( f(x)=\frac{10(1 - e^{x})}{x^{2}}\to-\infty \) (since numerator is negative and denominator positive, and both go to infinity, but numerator dominates in magnitude? Wait \( e^{x} \) grows faster than \( x^{2} \), so \( 1 - e^{x}\sim - e^{x} \), so \( f(x)\sim\frac{-10e^{x}}{x^{2}}\to-\infty \) as \( x\to\infty \).

For \( x\to-\infty \):

\( e^{x}\to0 \), so \( 1 - e^{x}\to1 \), denominator \( x^{2}\to\infty \), so \( f(x)=\frac{10(1 - e^{x})}{x^{2}}\to0^{+} \) (since positive numerator and positive denominator, numerator approaches 10, denominator approaches infinity).

Now, let's re - analyze the behavior:

- \( x < 0 \): \( f(x)>0 \) (above x - axis), as \( x\to0^{-} \), \( f(x)\to+\infty \).
- \( x > 0 \): \( f(x)<0 \) (below x - axis), as \( x\to0^{+} \), \( f(x)\to-\infty \).
- \( x\to\infty \): \( f(x)\to-\infty \).
- \( x\to-\infty \): \( f(x)\to0^{+} \).

Now, looking at the fourth graph (the one with the coordinate system):

- For \( x < 0 \) (left of y - axis), the graph is below the x - axis? No, the red curve on the left ( \( x < 0 \) ) has the arrow going down (towards \( - 2 \) ), but according to our analysis, it should be above x - axis and go to \( +\infty \) as \( x\to0^{-} \). Wait maybe the first three graphs are the ones with the symbols, and the fourth is the detailed one. Wait the user provided four graphs: first three are rectangles with some symbols, fourth is the one with x - y axes. Wait maybe the correct graph is the one where:

- Left side ( \( x < 0 \) ): above x - axis, going to \( +\infty \) near \( x = 0 \), going to \( 0^{+} \) as \( x\to-\infty \).
- Right side ( \( x > 0 \) ): below x - axis, going to \( -\infty \) near \( x = 0 \), going to \( -\infty \) as \( x\to\infty \).

Looking at the fourth graph, the left part ( \( x < 0 \) ) has the curve above x - axis? Wait the fourth graph's left curve ( \( x < 0 \) ) is below the x - axis (since the arrow is going down), no. Wait maybe I made a mistake in the function. Wait the function is \( f(x)=\frac{10 - 10e^{x}}{x^{2}}=10\frac{1 - e^{x}}{x^{2}} \). Let's compute \( f(-1)=\frac{10 - 10e^{-1}}{(-1)^{2}}=10(1 - \frac{1}{e})\approx10(1 - 0.3679)=10\times0.6321 = 6.321>0 \). \( f(1)=\frac{10 - 10e^{1}}{1^{2}}=10(1 - e)\approx10(1 - 2.718)= - 17.18<0 \). \( f(2)=\frac{10 - 10e^{2}}{4}=\frac{10(1 - e^{2})}{4}\approx\frac{10(1 - 7.389)}{4}=\frac{10(-6.389)}{4}\approx - 15.97<0 \). \( f(-2)=\frac{10 - 10e^{-2}}{4}=\frac{10(1 - \frac{1}{e^{2}})}{4}\approx\frac{10(1 - 0.1353)}{4}=\frac{10\times0.8647}{4}\approx2.16>0 \).

So at \( x=-2 \), \( f(-2)>0 \); \( x=-1 \), \( f(-1)>0 \); \( x = 1 \), \( f(1)<0 \); \( x = 2 \), \( f(2)<0 \).

Now, the fourth graph: let's see the x - axis from - 2 to 2, y - axis from - 2 to 2. The left curve ( \( x < 0 \) ) at \( x=-1 \) should be above x - axis (y>0), and at \( x=-2 \) also above x - axis (y>0). The right curve ( \( x > 0 \) ) at \( x = 1 \) should be below x - axis (y<0), at \( x = 2 \) below x - axis (y<0). The fourth graph's left curve ( \( x < 0 \) ) has the arrow going down (towards y = - 2), which would mean y<0 at \( x=-1 \), which contradicts \( f(-1)>0 \). Wait maybe the fourth graph is not the right one. Wait the first three graphs: the first graph has a symbol like a triangle, the second a cross, the third a plus - like symbol. Wait maybe the correct graph is the one where for \( x < 0 \), the function is positive (above x - axis) and for \( x > 0 \), negative (below x

Answer:

Step1: Analyze the function's domain

The function \( f(x)=\frac{10 - 10e^{x}}{x^{2}} \) has a denominator \( x^{2} \), so \( x
eq0 \). The graph should have a vertical asymptote at \( x = 0 \).

Step2: Analyze the function's behavior as \( x\to0^{+} \)

As \( x\to0^{+} \), \( e^{x}\to1 \), so numerator \( 10 - 10e^{x}\to0^{-} \), and \( x^{2}\to0^{+} \). So \( f(x)=\frac{\text{negative small}}{\text{positive small}}\to-\infty \).

Step3: Analyze the function's behavior as \( x\to0^{-} \)

As \( x\to0^{-} \), \( e^{x}\to1 \), numerator \( 10 - 10e^{x}\to0^{-} \), and \( x^{2}\to0^{+} \) (since square of negative is positive). So \( f(x)=\frac{\text{negative small}}{\text{positive small}}\to-\infty \).

Step4: Analyze the function's behavior as \( x\to\pm\infty \)

As \( x\to\pm\infty \), \( e^{x}\to\infty \) (for \( x\to\infty \)) or \( e^{x}\to0 \) (for \( x\to-\infty \)). For \( x\to\infty \): numerator \( 10 - 10e^{x}\to-\infty \), denominator \( x^{2}\to\infty \), so \( f(x)\to-\infty \). For \( x\to-\infty \): numerator \( 10 - 10e^{x}\to10 - 0 = 10 \), denominator \( x^{2}\to\infty \), so \( f(x)\to0^{+} \).

Now, looking at the given graphs, the fourth graph (the one with the coordinate system) has behavior: near \( x = 0 \), both sides go to \( -\infty \), as \( x\to\infty \) goes to \( -\infty \), as \( x\to-\infty \) goes to \( 0^{+} \)? Wait, no, maybe I made a mistake. Wait the fourth graph shown: the left and right parts (for \( x < 0 \) and \( x > 0 \)): for \( x > 0 \), as \( x\) approaches 0 from right, the graph goes down (to \( -\infty \)), and as \( x\to\infty \), goes up? Wait no, the red arrows: for \( x > 0 \), the graph is a U - like but wait no, the arrows: on the right side ( \( x>0 \) ), the arrow is going up, on the left side ( \( x < 0 \) ), arrow going down? Wait maybe I misanalyzed. Wait the function is \( f(x)=\frac{10 - 10e^{x}}{x^{2}}=10\frac{1 - e^{x}}{x^{2}} \). Let's re - evaluate:

Wait \( f(x)=\frac{10(1 - e^{x})}{x^{2}} \). For \( x = 1 \): \( f(1)=\frac{10(1 - e)}{1}\approx10(1 - 2.718)=10(-1.718)= - 17.18 \). For \( x=-1 \): \( f(-1)=\frac{10(1 - e^{-1})}{1}\approx10(1 - 0.368)=10(0.632)=6.32 \).

So at \( x = - 1 \), \( f(-1)>0 \); at \( x = 1 \), \( f(1)<0 \). Now, looking at the fourth graph: for \( x < 0 \) (left of y - axis), the graph is below or above? The fourth graph: the left part ( \( x < 0 \) ) has arrow going down (to \( - 2 \) direction) but wait no, the coordinate system: x from - 2 to 2, y from - 2 to 2. Wait maybe the fourth graph is not the right one. Wait the first three graphs are blank except for some symbols, the fourth has the coordinate. Wait maybe the correct graph is the one where for \( x>0 \), as \( x\to0^{+} \), \( f(x)\to-\infty \), \( x\to\infty \), \( f(x)\to-\infty \); for \( x < 0 \), \( x\to0^{-} \), \( f(x)\to-\infty \), \( x\to-\infty \), \( f(x)\to0^{+} \). The fourth graph (the one with the x - y axes) shows: for \( x>0 \), the graph is a U - shape? No, the red arrows: on the right side ( \( x>0 \) ), the graph goes up (arrow up), on the left side ( \( x < 0 \) ), arrow down. Wait maybe I messed up the function. Wait the function is \( f(x)=\frac{10 - 10e^{x}}{x^{2}}=10\frac{1 - e^{x}}{x^{2}} \). Let's take derivative to check monotonicity, but maybe the key is the vertical asymptote at \( x = 0 \), and the sign.

Wait when \( x>0 \): \( e^{x}>1 \), so \( 1 - e^{x}<0 \), so \( f(x)<0 \). When \( x < 0 \): \( e^{x}<1 \), so \( 1 - e^{x}>0 \), so \( f(x)>0 \). Ah! That's a key point. So for \( x>0 \), \( f(x)<0 \); for \( x < 0 \), \( f(x)>0 \). Now, looking at the fourth graph: the right part ( \( x>0 \) ) is below the x - axis (since \( f(x)<0 \)), and the left part ( \( x < 0 \) ) is above the x - axis (since \( f(x)>0 \)). Now, the fourth graph: the left part ( \( x < 0 \) ) has the arrow going down (but if \( f(x)>0 \) for \( x < 0 \), it should be above x - axis), and the right part ( \( x>0 \) ) is above? Wait no, the fourth graph's right part is above y = 0? Wait the fourth graph: the red curves: for \( x>0 \), the curve is above the x - axis (y>0), but we just found \( f(x)<0 \) for \( x>0 \). So maybe I made a mistake in sign. Wait \( f(x)=\frac{10 - 10e^{x}}{x^{2}}=10\frac{1 - e^{x}}{x^{2}} \). For \( x>0 \), \( e^{x}>1 \), so \( 1 - e^{x}<0 \), so \( f(x)<0 \) (correct). For \( x < 0 \), \( e^{x}<1 \), so \( 1 - e^{x}>0 \), so \( f(x)>0 \) (correct). So the graph should have: for \( x < 0 \), \( f(x)>0 \) (above x - axis), as \( x\to0^{-} \), \( f(x)\to-\infty \)? Wait no, earlier analysis: as \( x\to0^{-} \), numerator \( 10 - 10e^{x}\to0^{-} \) (since \( e^{x}\to1 \), so \( 10 - 10e^{x}\to0 \) from below), denominator \( x^{2}\to0^{+} \), so \( f(x)=\frac{\text{negative small}}{\text{positive small}}\to-\infty \). Wait that contradicts the sign from \( 1 - e^{x} \). Wait \( 1 - e^{x} \) when \( x\to0^{-} \): \( x\) is negative, \( e^{x}=e^{-|x|}=\frac{1}{e^{|x|}} \), so \( 1 - e^{x}=1 - \frac{1}{e^{|x|}} \). As \( |x|\to0 \), \( e^{|x|}\to1 \), so \( 1 - \frac{1}{e^{|x|}}\to0^{-} \) (since \( \frac{1}{e^{|x|}}>1 \) when \( |x| < 0 \)? No, \( |x| \) is non - negative, when \( x\to0^{-} \), \( |x|\to0^{+} \), \( e^{|x|}\to1^{+} \), so \( \frac{1}{e^{|x|}}\to1^{-} \), so \( 1 - \frac{1}{e^{|x|}}\to0^{+} \). Oh! I made a mistake earlier. \( e^{x} \) when \( x\to0^{-} \): \( x=-a \), \( a\to0^{+} \), \( e^{-a}=\frac{1}{e^{a}}\to1^{-} \), so \( 10 - 10e^{x}=10(1 - e^{-a})\to10(1 - 1^{+})=10(0^{-}) \)? No, \( 1 - e^{-a}=1-\frac{1}{e^{a}}=\frac{e^{a}-1}{e^{a}} \), as \( a\to0^{+} \), \( e^{a}-1\to0^{+} \), \( e^{a}\to1^{+} \), so \( 1 - e^{-a}\to0^{+} \). So numerator \( 10 - 10e^{x}=10(1 - e^{x})\to0^{+} \) when \( x\to0^{-} \). And denominator \( x^{2}\to0^{+} \). So \( f(x)=\frac{\text{positive small}}{\text{positive small}}\to+\infty \) when \( x\to0^{-} \). Ah! I had a sign error earlier. Let's re - do the limit:

For \( x\to0^{-} \):

\( e^{x}=e^{-|x|}=\frac{1}{e^{|x|}} \), so \( 1 - e^{x}=1-\frac{1}{e^{|x|}}=\frac{e^{|x|}-1}{e^{|x|}} \). As \( |x|\to0^{+} \), \( e^{|x|}-1\sim|x| \) (Taylor series), so \( 1 - e^{x}\sim\frac{|x|}{1}=|x|>0 \) (since \( |x|>0 \) when approaching 0 from left). So numerator \( 10(1 - e^{x})\sim10|x|>0 \), denominator \( x^{2}=|x|^{2}>0 \). So \( f(x)=\frac{10|x|}{|x|^{2}}=\frac{10}{|x|}\to+\infty \) as \( x\to0^{-} \) (since \( |x|\to0^{+} \)).

For \( x\to0^{+} \):

\( e^{x}\to1^{+} \), so \( 1 - e^{x}\to0^{-} \), numerator \( 10(1 - e^{x})\to0^{-} \), denominator \( x^{2}\to0^{+} \), so \( f(x)=\frac{10(1 - e^{x})}{x^{2}}\to-\infty \) (since negative over positive).

For \( x\to\infty \):

\( e^{x}\to\infty \), so \( 1 - e^{x}\to-\infty \), denominator \( x^{2}\to\infty \), so \( f(x)=\frac{10(1 - e^{x})}{x^{2}}\to-\infty \) (since numerator is negative and denominator positive, and both go to infinity, but numerator dominates in magnitude? Wait \( e^{x} \) grows faster than \( x^{2} \), so \( 1 - e^{x}\sim - e^{x} \), so \( f(x)\sim\frac{-10e^{x}}{x^{2}}\to-\infty \) as \( x\to\infty \).

For \( x\to-\infty \):

\( e^{x}\to0 \), so \( 1 - e^{x}\to1 \), denominator \( x^{2}\to\infty \), so \( f(x)=\frac{10(1 - e^{x})}{x^{2}}\to0^{+} \) (since positive numerator and positive denominator, numerator approaches 10, denominator approaches infinity).

Now, let's re - analyze the behavior:

• \( x < 0 \): \( f(x)>0 \) (above x - axis), as \( x\to0^{-} \), \( f(x)\to+\infty \).
• \( x > 0 \): \( f(x)<0 \) (below x - axis), as \( x\to0^{+} \), \( f(x)\to-\infty \).
• \( x\to\infty \): \( f(x)\to-\infty \).
• \( x\to-\infty \): \( f(x)\to0^{+} \).

Now, looking at the fourth graph (the one with the coordinate system):

• For \( x < 0 \) (left of y - axis), the graph is below the x - axis? No, the red curve on the left ( \( x < 0 \) ) has the arrow going down (towards \( - 2 \) ), but according to our analysis, it should be above x - axis and go to \( +\infty \) as \( x\to0^{-} \). Wait maybe the first three graphs are the ones with the symbols, and the fourth is the detailed one. Wait the user provided four graphs: first three are rectangles with some symbols, fourth is the one with x - y axes. Wait maybe the correct graph is the one where:

• Left side ( \( x < 0 \) ): above x - axis, going to \( +\infty \) near \( x = 0 \), going to \( 0^{+} \) as \( x\to-\infty \).
• Right side ( \( x > 0 \) ): below x - axis, going to \( -\infty \) near \( x = 0 \), going to \( -\infty \) as \( x\to\infty \).

Looking at the fourth graph, the left part ( \( x < 0 \) ) has the curve above x - axis? Wait the fourth graph's left curve ( \( x < 0 \) ) is below the x - axis (since the arrow is going down), no. Wait maybe I made a mistake in the function. Wait the function is \( f(x)=\frac{10 - 10e^{x}}{x^{2}}=10\frac{1 - e^{x}}{x^{2}} \). Let's compute \( f(-1)=\frac{10 - 10e^{-1}}{(-1)^{2}}=10(1 - \frac{1}{e})\approx10(1 - 0.3679)=10\times0.6321 = 6.321>0 \). \( f(1)=\frac{10 - 10e^{1}}{1^{2}}=10(1 - e)\approx10(1 - 2.718)= - 17.18<0 \). \( f(2)=\frac{10 - 10e^{2}}{4}=\frac{10(1 - e^{2})}{4}\approx\frac{10(1 - 7.389)}{4}=\frac{10(-6.389)}{4}\approx - 15.97<0 \). \( f(-2)=\frac{10 - 10e^{-2}}{4}=\frac{10(1 - \frac{1}{e^{2}})}{4}\approx\frac{10(1 - 0.1353)}{4}=\frac{10\times0.8647}{4}\approx2.16>0 \).

So at \( x=-2 \), \( f(-2)>0 \); \( x=-1 \), \( f(-1)>0 \); \( x = 1 \), \( f(1)<0 \); \( x = 2 \), \( f(2)<0 \).

Now, the fourth graph: let's see the x - axis from - 2 to 2, y - axis from - 2 to 2. The left curve ( \( x < 0 \) ) at \( x=-1 \) should be above x - axis (y>0), and at \( x=-2 \) also above x - axis (y>0). The right curve ( \( x > 0 \) ) at \( x = 1 \) should be below x - axis (y<0), at \( x = 2 \) below x - axis (y<0). The fourth graph's left curve ( \( x < 0 \) ) has the arrow going down (towards y = - 2), which would mean y<0 at \( x=-1 \), which contradicts \( f(-1)>0 \). Wait maybe the fourth graph is not the right one. Wait the first three graphs: the first graph has a symbol like a triangle, the second a cross, the third a plus - like symbol. Wait maybe the correct graph is the one where for \( x < 0 \), the function is positive (above x - axis) and for \( x > 0 \), negative (below x