Proof of transformations to find an approximate value

Question

I have no idea what this questions is asking, or how to go about solving.. can someone please help?

Answer:

Answer 1

The relationship "$f(x+h)\approx f(x) +hf'(x)$ for small h" is usually introduced long before the concept of Taylor series comes up. It is usually approached the first day of a calculus class as follows. Since we define $f'(x)$ as the limit of $\frac{f(x+h)-f(x)}{h}$ as $h\rightarrow 0$, we get an approximate value of $f'(x)$ by looking at the quotient for $h$ very close to $0$, that is, $f'(x)\approx \frac{f(x+h)-f(x)}{h}$ for small $h$. Multiplying both sides by $h$ and adding $f(x)$ to both sides, this gives the desired statement

$$f(x+h)\approx f(x) +hf'(x) \text{ for small }h$$

Here they are using $x=8$ and $h=0.06$. The formula becomes $f(8.06)\approx f(8) + 0.06f'(8)$, and using the function $f(x)=x^{1/3}$ (so that also $f'(x)=\frac13x^{-2/3}$) etc.