Let $l(x)$ be the linear approximation of $f(x) = x^{2/5}$ at $a = 32$. Approximation?

Question

I'm still a bit confused on how to figure out linear approximations. What are the basic steps to solving a problem like this? Thanks so much!

Let $l(x)$ be the linear approximation of $f(x) = x^{2/5}$ at $a = 32$

What is the linear approximation?

Thanks for the help!

Answer 1

The tangent line will be the linear approximation

ie,

$$f(x)=x^{\frac{2}{5}}$$

$$f'(x)=\frac{2}{5}x^{\frac{-3}{5}}$$

$$f(32)=(32)^{2/5}=4$$

$$f'(32)= \frac{1}{20}$$

and so the equation of the tangent line L(x) at x=32 is,

$$L(x)=4+\frac{1}{20}(x-32)$$