Question Video: Finding the Linear Approximation of a Root Function to Estimate a Root Number | Nagwa Question Video: Finding the Linear Approximation of a Root Function to Estimate a Root Number | Nagwa

# Question Video: Finding the Linear Approximation of a Root Function to Estimate a Root Number Mathematics • Higher Education

By finding the linear approximation of the function π(π₯) = β(π₯) at a suitable value of π₯, estimate the value of β(100.5).

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### Video Transcript

By finding the linear approximation of the function π of π₯ is equal to the square root of π₯ as a suitable value of π₯, estimate the value of the square root of 100.5.

The question is asking us to estimate the value of the square root of 100.5 by finding a linear approximation of the function π of π₯ is equal to the square root of π₯ at a suitable value for π₯. Letβs start by recalling what a linear approximation of a function π of π₯ at the point π₯ equals π is. If π is differentiable at π₯ is equal to π, then we can approximate our function π of π₯ near π₯ is equal to π by using the tangent line. We call this a linear approximation, πΏ of π₯. Itβs equal to π evaluated at π plus the first derivative of π of π times π₯ minus π. So how are we going to use this to estimate our value of the square root of 100.5?

Since weβre asked to do this by finding a linear approximation of the square root of π₯, we need to choose our value of π. Remember, our approximation will be more accurate the closer we are to π₯ is equal to π. So we should choose a value of π which gives us a value close to the square root of 100.5. If we choose π is equal to 100, then we get that π evaluated at π is the square root of 100. And this is close to the square root of 100.5. So weβll take this value of π. So to find our linear approximation, we need π evaluated at π and π prime evaluated at π. We already found π evaluated at π; itβs the square root of 100. So letβs now find the first derivative of π evaluated at π.

We need to differentiate π of π₯ is equal to root π₯. To do this, weβll start by rewriting π of π₯ by using our laws of exponents. Root π₯ is equal to π₯ to the power of one-half. We can then differentiate this by using the power rule for differentiation. We multiply by our exponent of π₯ and reduce this exponent by one. This gives us π prime of π₯ is equal to one-half times π₯ to the power of negative one-half. And again, we will rewrite this by using our laws of exponents. π₯ to the power of negative one-half is the same as dividing by the square root of π₯. So π prime of π₯ is equal to one divided by two root π₯.

Weβre now ready to find π prime of π. Since π is equal to 100, this is the first derivative of π at π₯ is equal to 100. So we substitute π₯ is equal to 100 into our expression for π prime of π₯. We get one divided by two root 100. And the square roots of 100 is 10. So π prime of 100 is equal to one divided by 20. So weβre now ready to find the linear approximation of our function π of π₯ is equal to the square root of π₯ at 100. We get πΏ of π₯ is equal to π of 100 plus π prime of 100 times π₯ minus 100. We know that π of 100 is the square root of 100 and π prime of 100 is one divided by 20. So πΏ of π₯ is equal to root 100 plus one over 20 times π₯ minus 100.

However, we can simplify this further. The square root of 100 is equal to 10. Next, we can distribute one twentieth over our parentheses. We get π₯ over 20 minus 100 over 20. And 100 over 20 is equal to five. Finally, 10 minus five is equal to five. So our linear approximation of the square root of π₯ at 100 is equal to five plus π₯ over 20. Weβre now ready to approximate the square root of 100.5 by using our linear approximation. The square root of 100.5 is approximately equal to πΏ evaluated at 100.5. Substituting π₯ is equal to 100.5 into our expression for πΏ of π₯, we get five plus 100.5 divided by 20, which if we calculate, we get 10.025.

Therefore, by finding the linear approximation of the function π of π₯ is equal to the square root of π₯ at π₯ is equal to 100, weβve shown the square root of 100.5 is approximately equal to 10.025.

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