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suppose $f(x) = 17 \\cdot 1.2^x + 65$. a) what is the domain of $f(x)$?…

Question

suppose $f(x) = 17 \cdot 1.2^x + 65$.
a) what is the domain of $f(x)$?
domain

b) what is the range of $f(x)$?
range

c) what is the $y$-intercept of $f(x)$? enter your answer as an ordered pair.
$y$-intercept

d) is $f(x)$ an increasing or decreasing function?
\\(\bigcirc\\) increasing
\\(\bigcirc\\) decreasing

e) what is the equation of the horizontal asymptote?

Explanation:

Part (a)

Step1: Analyze the function type

The function \( f(x) = 17\cdot1.2^{x}+65 \) is an exponential function. For exponential functions of the form \( a\cdot b^{x}+c \) (where \( a
eq0 \), \( b > 0,b
eq1 \)), the domain is all real numbers because we can raise \( b \) to any real power \( x \).

Step2: Determine the domain

Since there are no restrictions on the input \( x \) (no division by zero, no square roots of negative numbers, etc. that would limit \( x \)), the domain of \( f(x) \) is all real numbers. In interval notation, this is \( (-\infty,\infty) \).

Part (b)

Step1: Analyze the exponential term

For the exponential part \( 1.2^{x} \), since the base \( 1.2>1 \), the function \( y = 1.2^{x} \) has a range of \( (0,\infty) \). When we multiply by 17 (a positive number), the range of \( 17\cdot1.2^{x} \) is \( (0,\infty) \) (because multiplying a positive number by a number in \( (0,\infty) \) still gives a number in \( (0,\infty) \)).

Step2: Analyze the entire function

Now, we add 65 to \( 17\cdot1.2^{x} \). So, if \( 17\cdot1.2^{x}>0 \), then \( 17\cdot1.2^{x}+65>65 \). As \( x
ightarrow-\infty \), \( 1.2^{x}
ightarrow0 \), so \( 17\cdot1.2^{x}+65
ightarrow65 \), and as \( x
ightarrow\infty \), \( 17\cdot1.2^{x}+65
ightarrow\infty \). So the range of \( f(x) \) is \( (65,\infty) \).

Part (c)

Step1: Recall the definition of y - intercept

The \( y \)-intercept of a function is the point where \( x = 0 \). So we need to find \( f(0) \).

Step2: Calculate \( f(0) \)

Substitute \( x = 0 \) into the function: \( f(0)=17\cdot1.2^{0}+65 \). Since any non - zero number to the power of 0 is 1, \( 1.2^{0}=1 \). Then \( f(0)=17\times1 + 65=17 + 65 = 82 \). So the \( y \)-intercept is the ordered pair \( (0,82) \).

Part (d)

Answer:

Step1: Recall the horizontal asymptote of exponential functions

For an exponential function of the form \( y=a\cdot b^{x}+c \), the horizontal asymptote is the line \( y = c \). This is because as \( x
ightarrow-\infty \), \( b^{x}
ightarrow0 \) (when \( b > 1 \)), so \( y=a\cdot0 + c=c \).

Step2: Determine the horizontal asymptote

In our function \( f(x)=17\cdot1.2^{x}+65 \), \( c = 65 \). So the equation of the horizontal asymptote is \( y = 65 \).

Part (a) Answer:

The domain of \( f(x) \) is \( (-\infty,\infty) \) (or all real numbers).

Part (b) Answer:

The range of \( f(x) \) is \( (65,\infty) \).

Part (c) Answer:

The \( y \)-intercept is \( (0,82) \).

Part (d) Answer:

The function is increasing, so the answer is "Increasing".

Part (e) Answer:

The equation of the horizontal asymptote is \( y = 65 \).