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