Example 1: Tangent Line Approximation

Use linearization to approximate $\sqrt{4.14}$.

Step 1: Determine the $f\left(x\right)$ equation that you are trying to approximate.

Since we are trying to approximate the square root of a value, we will use the square root function as our $f\left(x\right)$.

$f\left(x\right)=\sqrt{x}$

Step 2: Determine the x-value that is closest to the actual x-value that you care about, which you will use to create your tangent line.

The thought process you will want to use to find the x-value is to ask yourself, “What is an x-value that is very, very, very close to the one I am trying to approximate , and that I could plug into the $f\left(x\right)$ and solve the math problem in my head?”

For this problem you would want an x-value very, very, very close to 4.14 that you could plug into the square root function, the $f\left(x\right)$, and do the math in your head.

That means you would want to choose $x_0=4$because 4 is very close to 4.14, and the$\sqrt{4}$ is a math problem you can do in your head.

$x_0=4$

Step 3: Find the equation of the tangent line to your $f\left(x\right)$  at the x-value you determined in Step 2 .

You will follow the exact same steps that you would to find the equation of any tangent line.

Step 1: Find the tangent point, $\left(x_1,y_1\right)$.

$\left(x_1,y_1\right)=(x_0,f(x_0\left)\right)$

Step 2: Find the slope , m , of the tangent line.

$m=f\prime \left(x_0\right)$

Step 3: Find the equation of the tangent line.

$y=m(x–x_1)+y_1$

The equation of the tangent line is your linearization equation, your tangent line approximation equation, or your local linear approximation equation, depending on who is asking. ????

Step 1:     $f\left(x\right)=\sqrt{x}, x_0=4$

$f\left(x_0\right)=\mathrm{ }f\left(4\right)=\sqrt{4}=2$

$(x_0,f(x_0\left)\right)=(4,2)$

$\left(x_1,y_1\right)= (4,2)$

Step 2:  $\mathrm{}$$f\left(x\right)=\sqrt{x}$

$f\left(x\right)=x^{\frac{1}{2}}$

$f\prime \left(x\right)={\frac{1}{2}x}^{–\frac{1}{2}}$

$f\prime \left(x_0\right)=f\prime \left(4\right)={\frac{1}{2}\left(4\right)}^{–\frac{1}{2}}=\frac{1}{4}$

$f\prime \left(x_0\right)=f\prime \left(4\right)=\frac{1}{4}$

$m=\frac{1}{4}$

Step 3: $\mathrm{}$$\left(x_1,y_1\right)= (4,2)$

$m=\frac{1}{4}$

$y=m(x–x_1)+y_1$

$y=\frac{1}{4}(x–4)+2$

Step 4: Use the linearization equation (tangent line equation) that you found in Step 3 to approximate the value you were initially asked to find.

What you just created is Step 3 is not the final answer it is now the equation that approximates the actual equation, $f\left(x\right)$ we care about, the square root function, $f\left(x\right)=\sqrt{x}$. It is the equation you use to find the final answer.

To approximate the value we are trying to find, $\sqrt{4.14}$, we plug 4.14 in for x in our approximation equation we just created in Step 3.

$\sqrt{x}\approx \frac{1}{4}(x–4)+2$

$\sqrt{4.14}\approx \frac{1}{4}(4.14–4)+2$

$\sqrt{4.14}\approx 2.035$

Final Result:

A linear equation that can provide you an approximation for the square root equation is $\sqrt{x}\approx \frac{1}{4}(x–4)+2$.

Depending on who is asking you that same equation could be represented as

–           Linearization Equation or Local Linear Approximation: $L\left(x\right)=\frac{1}{4}(x–4)+2$

–           Tangent Line Approximation Equation: $\hat{f}\left(x\right)=\frac{1}{4}(x–4)+2$

Plugging 4.14 in for x in our approximation equation tells us that $\sqrt{4.14}\approx 2.035$.