Wize AP Calculus (BC) Textbook > Applications of Differentiation: Contextual

Linearization (Linear Approximation)

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 Linearization  
Tangent Lines can give us great approximations to functions near a point. The process of using this linear approximation is called Linearization. 

Linearization
The linearization or linear approximation, of the functionf(x)f\left(x\right)f(x) about x=ax=ax=a is the functionL(x)L\left(x\right)L(x) defined by 
L(x)=f(a)+f′(a)(x−a)\boxed{L(x)=f(a)+f'(a)(x-a)}L(x)=f(a)+f′(a)(x−a)​
The value of aaa is sometimes called the center or the anchor point of the approximation. 

Note: The closer the input of L(x)L(x)L(x) is to aaa , the better the approximation of f(x)f(x)f(x) will be. 
Over/Under Estimates
If the function is concave up (upward U shaped in the neighborhood of the approximation) the approximation is an underestimate of the actual value.

If the function is concave down (downward U shaped in the neighborhood of the approximation) the approximation is an overestimate of the actual value.

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Example: Linearization
Approximate the square root of 26 without using a calculator.

Take f(x)=x ⇒f′(x)=12x\begin{aligned}

f(x)&=\sqrt{x}\\
\text{ }\\
\Rightarrow f'(x)&=\frac{1}{2\sqrt{x}}

\end{aligned}f(x) ⇒f′(x)​=x​=2x​1​​



Also, take a=25 ⇒f′(a)=1225=110\begin{aligned}

a&=25\\
\text{ }\\
\Rightarrow f'(a)&=\frac{1}{2\sqrt{25}}=\frac{1}{10}

\end{aligned}
a ⇒f′(a)​=25=225​1​=101​​



We have
f(x)≈f(x0)+f′(a)(x−a) ⇒f(26)=26≈f(25)+f′(25)[26−25]=25+110(1)=5+110 ⇒26≈5.1\begin{aligned}

f(x) &\approx f(x_0)+f'(a)(x-a)\\
\text{ }\\
\Rightarrow f(26)&=\sqrt{26}\approx f(25)+f'(25)[26-25]\\
&=\sqrt{25}+\frac{1}{10}(1)\\
&=5+\frac{1}{10}\\
\text{ }\\
\Rightarrow \sqrt{26}&\approx 5.1

\end{aligned}f(x) ⇒f(26) ⇒26​​≈f(x0​)+f′(a)(x−a)=26​≈f(25)+f′(25)[26−25]=25​+101​(1)=5+101​≈5.1​

Using the calculator
26≈5.0990195\sqrt{26}\approx 5.099019526​≈5.0990195

Use a linear approximation to estimate 8.012/38.01^{2/3}8.012/3

2+13002 + \dfrac{1}{300}2+3001​

4−13004 - \dfrac{1}{300}4−3001​

4+13004 + \dfrac{1}{300}4+3001​

2−13002 - \dfrac{1}{300}2−3001​

I don't know

Using a linear approximation, estimate g(1.1)g(1.1)g(1.1), given that  g(x)=f(x2+1)g(x)=f(x^2+1)g(x)=f(x2+1) and f(2)=f′(2)=1f(2)=f^{\prime}(2)=1f(2)=f′(2)=1

Answer

I don't know