Given the function f ( x ) = 4 x 2 − 1 f(x)=4x^2-1 f ( x ) = 4 x 2 − 1 , calculate the slope of the tangent line at x = − 3 x=-3 x = − 3 .

This problem says given the function f of x is equal to 4x squared minus 1, calculate the slope of the tangent line at x is equal to negative 3. Now the way we're going to solve this problem is by using the equation for the slope of a tangent line, which we've discussed is equal to the limit as x approaches c, c being some value we're given, and then the equation is f of x minus f of c all divided by x minus c. So this is the equation we need, and this is how we're going to solve the problem. Now what I'm going to go ahead and do is plug in the numbers that we have. Now I can see for c, that's going to be the value we're given, which is negative 3. So we're going to have the limit as x approaches negative 3. And then for our function, well, I can see that that's 4x squared minus 1. So we're gonna have 4x squared minus 1 as this first portion here, that's gonna be minus f of c. Now f of c is going to be our function with negative 3 plugged in. So what that's going to be is 4 times negative 3 squared minus 1. Now negative 3 squared, that's going to give you positive 9 because the negative signs will cancel when you multiply negative 3 by itself, then that's gonna be minus 1. Now 4 times 9, that's 36. We have 36 minus 1, which is 35. So that's going to be f of c. Now we're going to divide this by x minus c, which is x minus negative 3. And, of course, this can be rewritten as x plus 3. So what we end up having is our limit as x approaches negative 3, and we can go ahead and simplify what we have up here. So So what I'm going to do is I'm gonna take this negative one. I'm going to subtract off the 35, which will give me negative 36. So I have 4 x squared minus 36, and it's all going to be divided by x plus 3. I'm gonna go ahead and continue this on this right side of the page here. And what we can see is that we have a situation where we can try evaluating our limit. But if I go ahead and take this negative 3 and plug it in here, those will get negative 3 plus 3, which will give me a 0 on bottom. And that's something that we're not allowed to have in math, so we're going to need to factor. Now the way that I'm going to factor is by taking this 4 and factoring it out because I can see that 4, factoring out of the x squared is just going to give me x squared. And then factoring 4 out of 36, well, 4 divided by 36 is 9. Right? So we get x squared minus 9, and that's all going to be divided by x plus 3. Now minus 9 here, this 9 can also be written as 3 squared. And 3 squared, it would be a difference between 2 squares here, x squared minus 3 squared. So using this rule, we can recognize that this would be the 4 out in front, and then the difference of squares, x squared minus 3 squared would be x plus 3 times x minus 3. This is how it can factor if you go ahead and take this difference of squares, then divide that all by x plus 3. Now at this point, what I can do is cancel out the x plus threes. And so what I'm going to end up with is 4 times x minus 3. And keep in mind, this is something I actually forgot to keep consistent over here. We still have this limit being applied, and this is something you wanna keep track of because you don't wanna lose track of this limit right here. So you definitely wanna make sure that for each of these, you're including this limit as x approaches negative 3. But we still have this con consistent in this problem, so we have limit as x approaches negative 3. What I can do is take this negative 3 and plug it in for x. So that's going to give us 4 times negative 3 minus 3. Now negative 3 minus 3, that's negative 6, and 4 times negative 6 will give you negative 24. So this right here is going to be the solution for the slope of our tangent line at x is equal to negative 3 given the function that we have. So that way, there is the solution and the answer to this problem. So hope you found this video helpful. Thanks for watching, and please let me know if you have any questions.

2. Intro to Derivatives

Tangent Lines and Derivatives

Calculus

2. Intro to Derivatives

Tangent Lines and Derivatives

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Multiple Choice

Given the function $f(x)=4x^2-1$ , calculate the slope of the tangent line at $x=-3$ .

-36

-24

Verified step by step guidance

To find the slope of the tangent line to the function $ f(x) = 4x^2 - 1 $ at $ x = -3 $, we need to calculate the derivative of the function, $ f'(x) $. The derivative represents the slope of the tangent line at any point $ x $.

Start by differentiating $ f(x) = 4x^2 - 1 $. Use the power rule for differentiation, which states that if $ f(x) = ax^n $, then $ f'(x) = n \, ax^{n-1} $.

Apply the power rule to $ 4x^2 $: $ f'(x) = 2 \, \cdot \, 4x^{2-1} = 8x $. The derivative of a constant, $-1$, is $0$. Therefore, $ f'(x) = 8x $.

Now, substitute $ x = -3 $ into the derivative $ f'(x) = 8x $ to find the slope of the tangent line at this point: $ f'(-3) = 8 \, \cdot \, (-3) $.

Calculate $ f'(-3) $ to determine the slope of the tangent line at $ x = -3 $. This will give you the final slope value.

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Struggling with Calculus?

Given the function f(x)=4x2−1f(x)=4x^2-1f(x)=4x2−1, calculate the slope of the tangent line at x=−3x=-3x=−3.

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Tangent Lines and Derivatives practice set

Given the function $f(x)=4x^2-1$ , calculate the slope of the tangent line at $x=-3$ .