Use linear approximation, i.e. the tangent line, to approximate 5.3^2 as follows:
Let f(x)=x^2 and find the equation of the tangent line to f(x) at x=5 . Using this, find your approximation for 5.3^2

Question

Use linear approximation, i.e. the tangent line, to approximate 5.3^2  as follows: 
Let f(x)=x^2 and find the equation of the tangent line to f(x) at x=5 . Using this, find your approximation for 5.3^2

anonymous · Answer

anyone have a clue?

anonymous · Answer

$$
f(x)\approx L(x)
$$Where $$
L(x) = f'(a)(x-a)+f(a)
$$

anonymous · Answer

Typically $a$ is some point that is close to $x$ and $f(a)$ is easier calculated than $f(x)$ is.

anonymous · Answer

In this case $a=5$ and $x=5.3$.

anonymous · Answer

So first we find $f'(5)=2x|_{x=5}=2(5)=10$

anonymous · Answer

Next we find $f(5)=x^2|_{x=5}=(5)^2=25$

anonymous · Answer

This means $$
f(x)\approx L(x) = 10(x-5)+25 = 10x-50+25=10x-25
$$

anonymous · Answer

So given that $$
L(x)= 10x-15
$$We can approximate $f(5.3)$:$$
L(5.3)=10(5.3)-25=53-25=28
$$

anonymous · Answer

In short $f(5.3)\approx 28$