Question

1. consider

$f(x) = \frac{x^2 - 3x - 4}{x^2}$, $f(x) = \frac{3x + 8}{x^3}$, $f(x) = \frac{-6x - 24}{x^4}$

a. 1 mark find the x-intercept(s) of $f(x)$.

b. 5 marks determine where the function is increasing and decreasing. find the relative extreme value(s) of $f(x)$.

Explanation:

Step1: Find x-intercepts (set \( f(x) = 0 \))

To find the x - intercepts of the function \( f(x)=\frac{x^{2}-3x - 4}{x^{2}} \), we set the numerator equal to zero (since the denominator \( x^{2}
eq0 \) for the function to be defined) because a fraction is zero when its numerator is zero (and denominator is non - zero). So we solve the equation \( x^{2}-3x - 4=0 \).
We factor the quadratic equation: \( x^{2}-3x - 4=(x + 1)(x - 4)=0 \).
Using the zero - product property, if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \). So \( x+1 = 0 \) gives \( x=-1 \) and \( x - 4=0 \) gives \( x = 4 \).

Step2: Determine where the function is increasing/decreasing (use \( f'(x) \))

First, we are given \( f'(x)=\frac{3x + 8}{x^{3}} \). To find where the function is increasing or decreasing, we find the critical points by setting \( f'(x)=0 \) or finding where \( f'(x) \) is undefined.

• Set \( f'(x)=0 \):

Set the numerator \( 3x + 8=0 \). Solving for \( x \), we get \( 3x=-8\Rightarrow x=-\frac{8}{3} \).

• Find where \( f'(x) \) is undefined:

The denominator \( x^{3}=0 \) when \( x = 0 \). But \( x = 0 \) is not in the domain of \( f(x) \) (since \( f(x)=\frac{x^{2}-3x - 4}{x^{2}} \) has a denominator \( x^{2} \) which is zero at \( x = 0 \)).
Now we test intervals determined by the critical point \( x=-\frac{8}{3} \) and the point where the function is undefined \( x = 0 \). The intervals are \( (-\infty,-\frac{8}{3}) \), \( (-\frac{8}{3},0) \), and \( (0,\infty) \).

• For \( x\in(-\infty,-\frac{8}{3}) \), let's choose a test point, say \( x=-3 \). Then \( f'(-3)=\frac{3(-3)+8}{(-3)^{3}}=\frac{-9 + 8}{-27}=\frac{-1}{-27}=\frac{1}{27}>0 \). So the function is increasing on \( (-\infty,-\frac{8}{3}) \).
• For \( x\in(-\frac{8}{3},0) \), let's choose a test point, say \( x=-1 \). Then \( f'(-1)=\frac{3(-1)+8}{(-1)^{3}}=\frac{-3 + 8}{-1}=\frac{5}{-1}=-5<0 \). So the function is decreasing on \( (-\frac{8}{3},0) \).
• For \( x\in(0,\infty) \), let's choose a test point, say \( x = 1 \). Then \( f'(1)=\frac{3(1)+8}{(1)^{3}}=\frac{3 + 8}{1}=11>0 \). So the function is increasing on \( (0,\infty) \).

Step3: Find relative extreme values (use \( f'(x) \) and \( f''(x) \))

We are given \( f''(x)=\frac{-6x - 24}{x^{4}}=\frac{-6(x + 4)}{x^{4}} \).
First, we use the critical point from \( f'(x) \), which is \( x =-\frac{8}{3} \).
We substitute \( x =-\frac{8}{3} \) into \( f''(x) \):
\( f''(-\frac{8}{3})=\frac{-6(-\frac{8}{3}+4)}{(-\frac{8}{3})^{4}}=\frac{-6(\frac{-8 + 12}{3})}{(\frac{4096}{81})}=\frac{-6\times\frac{4}{3}}{\frac{4096}{81}}=\frac{-8}{\frac{4096}{81}}=-8\times\frac{81}{4096}=-\frac{81}{512}<0 \)
Since \( f''(-\frac{8}{3})<0 \), the function has a relative maximum at \( x =-\frac{8}{3} \).
We find the value of \( f(x) \) at \( x =-\frac{8}{3} \):
\( f(-\frac{8}{3})=\frac{(-\frac{8}{3})^{2}-3(-\frac{8}{3})-4}{(-\frac{8}{3})^{2}}=\frac{\frac{64}{9}+8 - 4}{\frac{64}{9}}=\frac{\frac{64}{9}+4}{\frac{64}{9}}=\frac{\frac{64 + 36}{9}}{\frac{64}{9}}=\frac{100}{64}=\frac{25}{16} \)

Answer:

• x - intercepts: \( x=-1 \) and \( x = 4 \)
• Increasing/Decreasing Intervals: Increasing on \( (-\infty,-\frac{8}{3})\cup(0,\infty) \), Decreasing on \( (-\frac{8}{3},0) \)
• Relative Extreme Value: Relative maximum at \( x =-\frac{8}{3} \) with \( f(-\frac{8}{3})=\frac{25}{16} \)