If f ( x ) = x + 12 f\left(x\right)=\sqrt{x}+12 , use the linearization L ( x ) L\left(x\right) at a = 16 a=16 to approximate f ( 16.01 ) f\left(16.01\right) .

Here, we are given a function, and we need to use linearization at an a value of 16 to approximate f of 16.01. Now recall, whenever we have values or situations where the a value is close to the value we're approximating, we can say that our function f of x is approximately equal to l of x. And in this equation, l of x is the linearization, which is going to be equal to f of a, our output value at our value of interest a, plus f prime of a times x minus a. This will give us the equation we're looking for, which is going to be the equation of a line that will help us to solve the problem. So let's go ahead and begin with finding this l of s here. Now what I'm first going to do is evaluate f of a. Now f of a is going to be f of 16 because you can see our a value is 16. Now all I need to do is take 16 and plug it in for x in this equation. We'll get the square root of 16 plus 12. Now doing this is going to give you 16. The square root of 16 is 4, plus 12, that's 16. So that's our f of a. Now we need next need to find f prime of a, which is gonna be a little bit trickier to find, which is f prime of 16. But to do this, I'm going to take the derivative of this whole equation right here. And the square root of x, that can also be written as x to the one half power, and it will have plus 12. Now taking this derivative here, I can go ahead and bring this power to the front and subtract it by 1. That's going to give me 1 half x to the minus one half power, then this 12 will just go away. So we can go ahead and simplify this by recognizing that this negative one half power will bring it to a positive one half power on the bottom, and x to the one half power is just the same thing as the square root of x. So this right here would be f prime of x, but we need to plug in our value of 16. So doing that, we're going to have 1 over 2 times the square root of 16, which is going to give us 1 over 8 as our result. So this right here would be the solution for f prime of 16. So we've gone ahead and we found our f prime of a, and we found our f of a value. So So since we've gone ahead and found these, what I need to do is go ahead and take these values and put them in to the equation we have over here. So doing this, we're going to have that l of x is equal to f of a, which we found to be 16, plus f prime of a, which we found to be 1 eighth multiplied by x minus our a value of 16. Now you can go ahead and simplify this equation. So we're going to have 16 plus, and then I'll distribute this 1 eighth into the parenthesis here. So we get 1 eighth of x minus and then 16 divided by 8 is 2, and then we'll have 16 and minus 2, which is 14. So it's gonna be 1 eighth x plus 14 as our linearization l of x. So now we've gone ahead and found our linearization, so we need to use this to approximate our value of interest at 16.01. So to do this, I'm gonna take this x. I'm going to replace it with 16.01 since this rule holds true when we have these really close values that we're looking at. So in this case, we'll have l of 16.01, and that's gonna be equal to 1 eighth of 16.01 plus 14. Now if you take 1 eighth of 16.01, this whole thing is gonna end up being 2.0125. It's very close to just being 2 because 16 divided by 8 we know is 2, but the point 1 does add on this decimal piece right here, then we add that to 14. And adding these two numbers will give you 16.0125. So So that right there is going to be the result and the solution to this problem. That's something that I'll say is if you were to actually take 16.01 and type it in for x and put that all into a calculator, you're going to find that it's very close to this approximation down here. And the reason for that is because these numbers are very close together, 1616.01. And that's the nice thing about linear approximation is the closer these numbers are together, the more accurate your approximation will be. But the farther and farther they are apart, the less accurate it will be. So as you can see, we got a pretty close approximation by using this process. Hope you found this video helpful.

4. Applications of Derivatives

Linearization

Calculus

4. Applications of Derivatives

Linearization

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Multiple Choice

If $f (x) = \sqrt{x} + 12$ , use the linearization $L (x)$ at $a equals 16$ to approximate $f of 16.01$ .

16.0125

16.0012

16.0000

5.2924

Verified step by step guidance

Identify the function f(x) = \sqrt{x} + 12 and the point a = 16 where we want to find the linearization.

Calculate the derivative f'(x) of the function f(x). The derivative of \sqrt{x} is \frac{1}{2\sqrt{x}}.

Evaluate the derivative at the point a = 16. This gives f'(16) = \frac{1}{2\sqrt{16}} = \frac{1}{8}.

Use the formula for linearization L(x) = f(a) + f'(a)(x - a). Substitute f(16) = \sqrt{16} + 12 = 16 and f'(16) = \frac{1}{8} into the formula.

Substitute x = 16.01 into the linearization L(x) to approximate f(16.01). Calculate L(16.01) = 16 + \frac{1}{8}(16.01 - 16).

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Struggling with Calculus?

If f(x)=x+12f\left(x\right)=\sqrt{x}+12, use the linearization L(x)L\left(x\right) at a=16a=16 to approximate f(16.01)f\left(16.01\right).

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Linearization practice set

If $f (x) = \sqrt{x} + 12$ , use the linearization $L (x)$ at $a equals 16$ to approximate $f of 16.01$ .