Question

1. (4 points) using the information given below,

a. (1 point) let g(x)=(f(x))³. find g(4).
b. (1 point) let h(x)=√xe^f(x). find h(4).
c. (2 points) find k(1) if k(x)=f⁻¹(x).
f(x)
(2,5)
p
(4,1)
tangent line

Explanation:

Step1: Find $g'(x)$ using chain - rule

The chain - rule states that if $g(x)=(f(x))^{3}$, then $g'(x) = 3(f(x))^{2}\cdot f'(x)$. We know from the graph that $f(4)=1$ and the slope of the tangent line to $y = f(x)$ at $x = 4$ gives $f'(4)$. The slope of the tangent line at the point $(4,1)$ can be found using the formula for the slope of a line. Since the tangent line passes through $(4,1)$ and we can assume another point on the tangent line (from the graph's linear - like behavior), the slope of the tangent line $m=f'(4)$. Let's assume the tangent line is linear near $x = 4$. The slope of the tangent line at $x = 4$ is some value. Let's say from the graph, if we consider the tangent - line approximation, $f'(4)$ is the slope of the line passing through $(4,1)$. Substituting $x = 4$ into $g'(x)$, we have $g'(4)=3(f(4))^{2}\cdot f'(4)$. Since $f(4)=1$, then $g'(4)=3\cdot1^{2}\cdot f'(4)=3f'(4)$. If we assume the slope of the tangent line at $(4,1)$ is $m = f'(4)=- 2$ (by observing the graph's trend), then $g'(4)=3\times(-2)=-6$.

Step2: Find $h'(x)$ using product - rule and chain - rule

The product - rule states that if $h(x)=\sqrt{x}e^{f(x)}=x^{\frac{1}{2}}e^{f(x)}$, then $h'(x)=\frac{1}{2}x^{-\frac{1}{2}}e^{f(x)}+x^{\frac{1}{2}}e^{f(x)}f'(x)$. Substitute $x = 4$: $h'(4)=\frac{1}{2}\cdot4^{-\frac{1}{2}}e^{f(4)}+4^{\frac{1}{2}}e^{f(4)}f'(4)$. Since $f(4)=1$, we have $h'(4)=\frac{1}{2}\cdot\frac{1}{2}e^{1}+2e^{1}f'(4)=\frac{1}{4}e + 2ef'(4)$. If $f'(4)=-2$, then $h'(4)=\frac{e}{4}+2e\times(-2)=\frac{e}{4}-4e=-\frac{15e}{4}$.

Step3: Find $k'(x)$ using the formula for the derivative of an inverse function

The formula for the derivative of an inverse function is $k'(x)=\frac{1}{f'(k(x))}$, where $k(x)=f^{-1}(x)$. We want to find $k'(1)$. Since $f(4) = 1$, then $k(1)=4$. So $k'(1)=\frac{1}{f'(k(1))}=\frac{1}{f'(4)}$. If $f'(4)=-2$, then $k'(1)=-\frac{1}{2}$.

Answer:

a. $g'(4)=-6$
b. $h'(4)=-\frac{15e}{4}$
c. $k'(1)=-\frac{1}{2}$