If f ( x ) = x 3 + 1 f\left(x\right)=x^3+1 f ( x ) = x 3 + 1 , use the linearization L ( x ) L\left(x\right) L ( x ) at a = 5 a=5 a = 5 to approximate f ( 5.1 ) f\left(5.1\right) f ( 5.1 ) .

Given the function f of x is equal to x cubed plus 1, use the linearization l of x at a equals 5 to approximate f of 5.1. Okay. Now to solve this problem, recall that the equation we need is this 1. L of x via linearization is f of a plus f prime of a times x minus a. This is the equation for the line that allows us to do these approximations that for these linear approximations for these values that are really close to our value of a. Since we can see 5.15 are very close values together, we can tell that our function is going to approximately be equal to the linearization at that specific point. So let's go ahead and see if we can find this linearization. Well, to do this, I'm first going to set this up by recognizing the first thing we need to find is f of a. So f of a, in this case, our a value is 5. So what I need to do is take every place I see x in this equation and replace it with 5. So we're gonna get 5 cubed plus 1. 5 cubed is a 125, and a 125 plus 1 is a 126. Now the next thing we need to find is f prime of a. F prime of a is going to be the derivative at this value right here, at our our interest value of a. So it's gonna be the derivative evaluated at 5. Now first, I'll use the power rule to find the derivative of this function, which is going to give us 3 x squared. That's gonna be and then this one is just gonna drive off. That's a constant, so we'll just end up with 3 x squared. Then pulling our value of 5 in, we'll get 3 times 5 squared. 5 squared is 25, and 25 times 3 is 75. So this is going to be our f prime of a value. Then that's all gonna be multiplied by x minus our value of a, which is 5. So this right here is going to be the linearization. And what I can do is try to simplify this to get it into the standard line equation. Now you can do this by the steps where you just go ahead and take the 75 and distribute it in here and then simplify the values you see. If you do this correctly, you should end up end up with l of x is equal to 75 x minus 249. That's what things should simplify to once you've gone through these steps. So this is the equation of our line right here. So now that we have the linearization, we can go ahead and use this to approximate f of 5.1. So we can approximate this by taking 5.1 and plugging it in to our linearization. So I'm gonna do that down here. So I have l of 5.1, which is going to be equal to, in this case, we're going to have 75 times 5.1, and that's gonna be minus 249. Now at this point, we just need to do the math. 75 times 5.1, that's going to come out to 382 point 5. So 382 and then plus 1 half, because if you go ahead and split this into 5 into point 1, you can distribute each one individually and add them, and you get 382.5, and you can subtract off 249. So from here, you just need to subtract these two numbers, which should come out to a 133 0.5. So this right here would be the linear approximation at a value of 5.1, and that would be the solution to this problem. Hope you found this video helpful.

4. Applications of Derivatives

Linearization

Calculus

4. Applications of Derivatives

Linearization

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Multiple Choice

If $f\left(x\right)=x^3+1$ , use the linearization $L\left(x\right)$ at $a=5$ to approximate $f\left(5.1\right)$ .

501.0

133.7

133.5

126.1

Verified step by step guidance

Identify the function f(x) = x^3 + 1 and the point a = 5 where we want to find the linearization.

Calculate the derivative of the function, f'(x) = 3x^2, to find the slope of the tangent line at x = a.

Evaluate the derivative at the point a = 5: f'(5) = 3(5)^2.

Use the formula for linearization L(x) = f(a) + f'(a)(x - a) to find the linear approximation. Substitute f(5) and f'(5) into the formula.

Substitute x = 5.1 into the linearization L(x) to approximate f(5.1).

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Struggling with Calculus?

If f(x)=x3+1f\left(x\right)=x^3+1f(x)=x3+1, use the linearization L(x)L\left(x\right)L(x) at a=5a=5a=5 to approximate f(5.1)f\left(5.1\right)f(5.1).

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Linearization practice set

If $f\left(x\right)=x^3+1$ , use the linearization $L\left(x\right)$ at $a=5$ to approximate $f\left(5.1\right)$ .