Finding Functions from Derivatives

In Exercis...

Question

Finding Functions from Derivatives

In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.

f'(x) = 2x − 1, P(0,0)

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the function that passes through the point P1 -4 and has the derivative of F of X to be 2 X minus 5. We want to use the second corollary of the mean value theorem to help us. A says F of X is X2 minus 5X. B says it's X2 + 5X, C X2 minus 5X + 10, and DX2 + 5 X minus 10. Now, how is the second corollary of the mean value theorem going to help us to figure out the function that passes through this point? Well, recall that this corollary basically tells us that if the derivative of FX is equal to the derivative of geofi, another function. At each point X. In the open interval AB. Then, OK, there exists a constant sea. Such that, OK. F of X is equal to G X plus C. For all values of X in the open interval. OK. So that's for all X in the open interval AB. In other words, F minus G is a constant function on A B. So now if we're going to figure out our function that passes through this point, first we'll need to find the value of GFX, OK? And then we need to substitute it into FFX equals GFX plus C to solve for for C and thus solve for FFX. Now remember, we were told that the derivative of FF X E equals the derivative of G of X, OK? So since the derivative of GFX. Is equal to. To X minus 5 based on our problem, then we have to ask ourselves what can we differentiate to get to X minus 5. Well, that means geoff X would have to be equal to X2 minus 5. Because the 5 X sorry, because the derivative of X2 is 2 X and the derivative of negative 5 X is -5. Thus this would imply then that F of X. Is going to be equal to x2 minus 5 X plus C. Now, since we're finding the function that passes through the point P1 -4, OK, then that means, OK, that X will be equal to 1 and F of X will be equal to -4. So if we substitute that into our equation for FX, that means -4 equals 12 minus 5 multiplied by 1 plus C. 1 squared is 15 multiplied by 1 is 5, and 1 minus 41 minus 5 equals -4. Therefore, -4 equals -4 plus C, which would imply then that C must be equal to 0. If C equals 0, this means that F of X will be equal to G of X X2 minus 5 X plus 0, or that tells us F X equals X2 minus 5 X. Therefore, the function that passes through the point P 14 and has a derivative of 2X minus 5 is X2 minus 5X. If we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helps.

Finding Functions from Derivatives

In Exercises 37–40, find the function with the given derivative whose graph passes through the point P.

f'(x) = 2x − 1, P(0,0)

Key Concepts

Antiderivatives

Integration

Initial Conditions

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