Question

f(x)=x^{3}-5x - 2, and f(3)=10. let g(x) be the inverse function of f(x). g(10)= write the equation of the tangent line to g(x) at x = 10. y=

Explanation:

Step1: Recall the formula for the derivative of an inverse function

If $g(x)$ is the inverse of $f(x)$, then $g^{\prime}(y)=\frac{1}{f^{\prime}(x)}$, where $y = f(x)$. Given $f(3)=10$, so when $y = 10$, $x = 3$. First, find the derivative of $f(x)$.
$f(x)=x^{3}-5x - 2$, then $f^{\prime}(x)=3x^{2}-5$.

Step2: Evaluate $f^{\prime}(3)$

Substitute $x = 3$ into $f^{\prime}(x)$.
$f^{\prime}(3)=3\times3^{2}-5=3\times9 - 5=27 - 5=22$.

Step3: Find $g^{\prime}(10)$

Since $g^{\prime}(10)=\frac{1}{f^{\prime}(3)}$, and $f^{\prime}(3)=22$, then $g^{\prime}(10)=\frac{1}{22}$.

Step4: Find the point on $g(x)$

Since $f(3)=10$, then $g(10)=3$. The point on the graph of $y = g(x)$ is $(10,3)$.

Step5: Write the equation of the tangent - line

The equation of a line in point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(10,3)$ and $m = g^{\prime}(10)=\frac{1}{22}$.
$y-3=\frac{1}{22}(x - 10)$.
Expand to get $y=\frac{1}{22}x-\frac{10}{22}+3=\frac{1}{22}x-\frac{5}{11}+3=\frac{1}{22}x+\frac{- 5 + 33}{11}=\frac{1}{22}x+\frac{28}{11}$.

Answer:

$g^{\prime}(10)=\frac{1}{22}$
$y=\frac{1}{22}x+\frac{28}{11}$