Given the function f ( x ) = x 2 + 100 f(x)=x^2+100 , calculate the slope of the tangent line at x = 0 x=0 .

Let's give this problem a try. So here we're told, given the function f of x is equal to x squared plus a 100, calculate the slope of the tangent line at x is equal to 0. Now in solving these types of problems, there's really only one equation you need to know, and that's that the slope of the tangent line is equal to the limit as x approaches some given value c of f of x minus f of c, all divided by x minus c. So this is the equation we need. Now what I can do is apply this equation to what we have in this problem, the function and the numbers that we're given. So what I'm gonna start by doing is figuring out what c is, and c is going to be this number that we have on that we're looking at on the x axis. So we have x is equal to 0. So that means that the slope of our tangent line is going to be the limit as x approaches 0. Now what we're going to have is the function, which I can see is x squared plus a100, and then that's going to be the function minus f of c, which would be f of 0. Now f of 0 is going to be we'll replace this x with 0, so we'll have 0 squared plus a100. And that's all gonna be equal to a100. So going to have this whole thing minus a100 and then all that is going to be divided by x minus and then c, which is 0. Now at this point, I can simplify some things. What I notice is that actually this positive a100 and negative a100 are going to cancel each other. So all we're going to end up with on top, we'll still have this limit continuing by the way, so we'll have the limit as x approaches 0. On top we're just gonna have x squared, and then on bottom we're going to have x minus 0, which is just x. Now at this point what I could try doing is plugging in this limit for 0, but notice we're gonna end up with a 0 on bottom. We cannot divide by 0. But what I can do is actually some pretty some pretty basic simplification here, because I notice that one of these x's is gonna cancel with these x on bottom. So all we're really going to have is the limit as x approaches 0 of x. And, well, from here, I can just go ahead and take this 0 here, and I can plug it in for x. So what we're going to have is that the slope of our tangent line is equal to, replacing this x with a 0, 0. And that's it. That right there is the solution. That means the slope of our tangent line is 0. What exactly does it mean to have 0 as your slope? Well, a 0 slope just means that we have some kind of straight line going on here. So our tangent line is looking something like this. It's just going straight across. And this actually makes sense, because if you think about it, what we have is we have a parabola in this problem. X squared plus a 100 is just some parabola that's been shifted a 100 units up. So if you think about a graph where we have a y axis and an x axis, you can think of this parabola as just being something that is shifted way up here, way up on the y axis, so it looks something like that. And we're looking at where the x axis is at 0, that's gonna be at this point right here or there on the parabola, it's not gonna have any slope. So it makes sense that this tangent line would have a slope of 0. Because if we go here, if you notice, it's just it's just gonna be a straight line going right across that point. So it makes sense that this slope would be 0, and it makes sense that we get the tangent line slope being 0. So that's what this really all means, but either way, this here is the solution to the problem, and that's how you can solve these types of problems. So hope you found this video helpful.

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23. Intro to Derivatives & Area Under the Curve

Tangent Lines & Derivatives

Precalculus

23. Intro to Derivatives & Area Under the Curve

Tangent Lines & Derivatives

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Multiple Choice

Given the function $f (x) = x^{2} + 100$ , calculate the slope of the tangent line at $x equals 0$ .

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To find the slope of the tangent line to the function $ f(x) = x^2 + 100 $ at $ x = 0 $, we need to calculate the derivative of the function, $ f'(x) $.

The derivative of $ f(x) = x^2 + 100 $ is found using the power rule. The power rule states that if $ f(x) = x^n $, then $ f'(x) = n \cdot x^{n-1} $.

Apply the power rule to $ x^2 $: $ \frac{d}{dx}(x^2) = 2x $. The derivative of a constant, such as 100, is 0.

Combine the derivatives: $ f'(x) = 2x + 0 = 2x $.

Evaluate the derivative at $ x = 0 $ to find the slope of the tangent line: $ f'(0) = 2 \cdot 0 = 0 $.

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Struggling with Precalculus?

Given the function f(x)=x2+100f(x)=x^2+100, calculate the slope of the tangent line at x=0x=0.

Watch next

Tangent Lines & Derivatives practice set

Given the function $f (x) = x^{2} + 100$ , calculate the slope of the tangent line at $x equals 0$ .