Use Linear Approximation to estimate $\Delta f=f(3.02)-f(3)$

Question

I have a homework question that is asking to use the linear approximation to estimate $\Delta f=f(3.02)-f(3)$ for $f(x)=x^2$.

I know that the linear approximation is $L(x) = f'(a)(x-a)+f(a)$.

I however do not quit understand what the problem is asking in the first place. Couldn't I just estimate $\Delta f=f(3.02)-f(3)$, by plugging both 3.02 and 3 into the original function of $f(x) = x^2$ and subtracting them. How do I do this with linear approximation?

Answer 1

Using the linear approximation $L$ for $f$, we have $$\Delta f := f(x) - f(a) \approx L(x) - L(a) = L(x) - f(a) = f'(a) (x - a).$$ (Notice that the last equality is just the rearrangement of the linear approximation formula $$L(x) = f'(a) (x - a) + f(a).)$$

In your case, $$f'(3) = \left.2x \right\vert_{x = 3} = 2(3) = 6,$$ and hence substituting in the above formula gives $$\Delta f = f(3.02) - f(3) \approx (6)((3.02) - 3) = 6(0.02) = \boxed{0.12}.$$

Answer 2

Hint: $$\begin{align} \frac{\Delta f}{0.02}&=\frac{f(3.02)-f(3)}{0.02}\\ &=\frac{f(3+0.02)-f(3)}{0.02}\\ &\approx \lim_{h\to0}\frac{f(3+h)-f(3)}{h}\\ &=f'(3) \end{align}$$

Answer 3

You can indeed compute: $$f(3.02)-f(3) = (3.02)^2-3^2 = (6.02)\cdot 0.02 = 0.1204,$$ the exact value. The point of the exercise is avoiding this computations. Using linear approximation gives: $$L(3.02) = f(3) + f'(3)\cdot0.02,$$ so that $\Delta f \approx f'(3)\cdot 0.02 = 6 \cdot 0.02 = 0.12$. Good, no?

Answer 4

here is what you can do to approximate $3.02^2 - 3^2$ using difference of squares:

$$3.02^2 - 3^2 = (3.02 - 3)(3.02 + 3) \simeq 0.02*6 = 0.12 $$