Q. 21 - Calculus

Question

Suppose that you know that the derivative of $f\left(x\right)=\sqrt{x}$ is equal to $f\text{'}\left(x\right)=\frac{1}{2\sqrt{x}}$. Use this fact, local linearity, and the fact that $\sqrt{4} = 2$ to approximate the value of $\sqrt{4.1}$ How close is your approximation with the approximation of $\sqrt{4.1}$ that you can find with a calculator? (Hint: Consider the tangent line to f(x) at x = 4, and use it to approximate the function nearby.) 

Step-by-Step Solution

Answer

$2.025$

Step 1. Given information

We have been given that that the derivative of $f\left(x\right)=\sqrt{x}$ is equal to $f\text{'}\left(x\right)=\frac{1}{2\sqrt{x}}$.

We have to approximate the value of $\sqrt{4.1}$ using the given fact, local linearity, and the fact that $\sqrt{4}=2$. How close is our approximation with the approximation of $\sqrt{4.1}$ that can be found with a calculator.

Step 2. Find the equation of the tangent line

To approximate the given value follow the steps :

Given : 

$$\begin{gathered} f\left(x\right)=\sqrt{x} \\ {f}^{\text{'}}\left(x\right)=\frac{1}{2\sqrt{x}} \end{gathered}$$

The slope of the tangent line at $x=4$ will be :

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(4\right)=\frac{1}{2\sqrt{4}}\\ =\frac{1}{2\cdot 2}\\ =\frac{1}{4}\end{array}$

Therefore, the equation of the tangent line at (4,2) with slope $\frac{1}{4}$ is:

$\begin{array}{r}y-2=\frac{1}{4}(x-4)\\ y-2=\frac{1}{4}x-1\\ y=\frac{1}{4}x+1\end{array}$

Step 3. Approximate the value

To approximate the value, replace x=4.1 in the equation $y=\frac{1}{4}x+1$

$\begin{array}{r}y=\frac{1}{4}\left(4.1\right)+1\\ =1.025+1\\ =2.025\end{array}$