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

give this problem a try. So here we're told given the function f of x is equal to 3 times x squared minus 1, find the equation of the tangent line at x is equal to 1. Now to solve this problem, my first mission here is going to be to find the slope of our tangent line. And to do that, we need this equation. It's gonna be the limit as x approaches our given value, and that's going to be f of x minus f of c all divided by x minus c. This is the equation we need. Now what I'm going to do is plug in the values that we have. First off, I need to find c, and c is gonna be this given x value that we're approaching. So we're going to have the limit as x approaches 1, and then this is going to be f of x, which is which we can see as 3 times x squared minus 1, and then that's going to be minus f of c. Now f of c is going to be f of our given x value here, 1, plugged into this function. So we're going to replace this x here with a 1. So we're gonna have 3 times 1 squared minus 1. One squared is just 1, so really we just have 1 minus 1, which is 0. So we have 3 times 0, which is 0. So that means we just need to plug 0 in there. Now this is all going to be divided by x minus c, where c is 1. Now at this point, what I can do is I can try to simplify what we have on top. So we'll have the limit as x approaches 1, and then we're going to have 3 times x squared minus 1 all over x minus 1. Now at this point, I can try to evaluate this limit. But notice if I plug this x value of 1 in for x, we'll get 1 minus 1. That's dividing by 0. We're not allowed to do that in math, but what I can try to do instead is factor the top of this equation to see if I can get that x minus 1 to cancel on bottom. Now what I notice is if we have x minus 1 or x squared minus 1, this would be equivalent to x squared minus 1 squared. And if I do this, I would have a difference in squares. So I can use the rule for difference in squares to factor this. So I have the limit as x approaches 1, and then we're going to have this 3 times. And then using the difference of squares, we'll have x plus one times x minus 1. This is always how you can factor these types of situations where you have a difference between 2 squares, and all of that's going to be divided by x minus 1. Now at this point, what I can do is I can cancel the x minus ones. So all we're going to have is 3 times x plus one left over. And at this point, I'm just to kinda save some room here, I'm gonna do this all in one step where I take this one and I plug it in for x because now we have this limit factor. So using the slope of the tangent line, I'll write it over here. What we're going to have is 3 times and then replacing this x with a one, we're going to have 3 times 1+1, which is 3 times 2, which is equal to 6. So this right here is the slope of our tangent line. Now oftentimes when we solve these types of problems where we're trying to find the slope, this would be our final answer. But we're not asked to just find the slope of the tangent line. We're asked to find the equation of the tangent line. And to do this, well, the standard point slope form is going to be y minus y naught or y y of 0 with this little zero subscript here. That's going to be equal m, the slope, times x minus x naught. Now we've gone ahead and calculated the slope already, so that's covered. But what I need to do is I need to find a point to plug in here so I can take out this y and this x naught here. And the way that I can do that is, well, I can just use a point from this original function. And the point I'm gonna use from this original function is with this given value of 1. So we already figured out that f of 1 is equal to 0, so that gives me the point where x is 0, y is equal or excuse me, where x is 1, y is equal to 0. So this is the point that we have. And this right here would be our x naught and our y naught, so we can go ahead and plug that in to this equation right here. So we have y minus y naught, which is 0, is equal to m, our slope of 6, times x minus x naught, which is 1. Now y minus 0 is just y, and what I can do is distribute the 6 into these parentheses. So we're gonna have 6x minus 6. And this right here is the equation of our tangent line. So as you can see, we get this standard form here, and this standard form is actually in the form y equals mx plus b. This is an equation that we've seen in the past for any kind of line and makes sense. M is our slope of 6 and b is our y intercept of negative 6. And as you can see, since we got that form, we did get the equation for a line, and this would be the tangent line at x equals 1. So that is how you can solve these types of problems. Hope you found this video helpful.

2. Intro to Derivatives

Tangent Lines and Derivatives

Calculus

2. Intro to Derivatives

Tangent Lines and Derivatives

Struggling with Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

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

$y=-6x+6$

$y=-6x-6$

$y=6x+6$

$y=6x-6$

Verified step by step guidance

First, find the derivative of the function f(x) = 3(x^2 - 1). The derivative, f'(x), represents the slope of the tangent line at any point x.

To find f'(x), apply the power rule to differentiate 3(x^2 - 1). The power rule states that d/dx [x^n] = n*x^(n-1).

Differentiate each term separately: the derivative of 3x^2 is 6x, and the derivative of -3 is 0. Therefore, f'(x) = 6x.

Evaluate the derivative at x = 1 to find the slope of the tangent line at this point. Substitute x = 1 into f'(x) to get f'(1) = 6*1 = 6.

Use the point-slope form of the equation of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point of tangency. Here, m = 6 and the point is (1, f(1)). Calculate f(1) = 3(1^2 - 1) = 0, so the equation becomes y - 0 = 6(x - 1). Simplify to get y = 6x - 6.