OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

anyone have a clue?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

In short $$f(5.3)\approx 28$$

5 years ago