Example 2: Tangent Line Approximation

Find a linear expression that approximates $f\left(x\right)=\sqrt{{5x}^4+{7x}^{–3}–10x}$ around $x=1$.

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 $f\left(x\right)$.

$f\left(x\right)=\sqrt{{5x}^4+{7x}^{–3}–10x}$

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.

In this example they are asking you to only find the linear expression, and not to use that expression to approximate a given value. That means that they are giving you the x-value that you need to use in your linear approximation equation (equation of the tangent line).

$x_0=1$

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:    
$f\left(x\right)=\sqrt{{5x}^4+{7x}^{–3}–10x}, x_0=1$  

$f\left(x_0\right)=\mathrm{ }f\left(1\right)=\sqrt{{5\left(1\right)}^4+{7\left(1\right)}^{–3}–10\left(1\right)}=\sqrt{2}$

$(x_0,f(x_0\left)\right)=(1,\sqrt{2})$

$\left(x_1,y_1\right)= (1,\sqrt{2})$

Step 2:  $\mathrm{}$$f\left(x\right)=\sqrt{{5x}^4+{7x}^{–3}–10x}$

$f\left(x\right)={({5x}^4+{7x}^{–3}–10x)}^{\frac{1}{2}}$

$f\prime \left(x\right)=({20x}^3–21x^{–4}–10)\frac{1}{2}{({5x}^4+{7x}^{–3}–10x)}^{–\frac{1}{2}}$

$f\prime \left(x_0\right)=f\prime \left(1\right)=({20\left(1\right)}^3–21{\left(1\right)}^{–4}–10)\frac{1}{2}{({5\left(1\right)}^4+{7\left(1\right)}^{–3}–10(1\left)\right)}^{–\frac{1}{2}}=\frac{–11}{2\sqrt{2}}$

$f\prime \left(x_0\right)=f\prime \left(1\right)=\frac{–11}{2\sqrt{2}}$

$m=\frac{–11}{2\sqrt{2}}$

Step 3: $\mathrm{}$$\left(x_1,y_1\right)= (1,\sqrt{2})$

$m=\frac{–11}{2\sqrt{2}}$

$y=m(x–x_1)+y_1$

$y=\frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}$

Step 4: Use the linearization equation (tangent line equation) that you found in Step 3 to as the linear expression that approximates the actual equation, $f\left(x\right)$ around x=1 .

$\sqrt{{5x}^4+{7x}^{–3}–10x}\approx \frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}$

$\sqrt{{5x}^4+{7x}^{–3}–10x}\approx \frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}$

Final Result:

A linear equation that can provide you an approximation for $\sqrt{{5x}^4+{7x}^{–3}–10x}\mathit{ is }\frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}.$

$\sqrt{{5x}^4+{7x}^{–3}–10x}\approx \frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}$

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

–           Linearization Equation or Local Linear Approximation: $L\left(x\right)=\frac{–11}{2\sqrt{2}}(x–1)+\sqrt{2}$

–           Tangent Line Approximation Equation: $\hat{f}\left(x\right)=\frac{1}{4}\frac{–11}{2\sqrt{2}}\left(x–1\right)+\sqrt{2}$