`f'(x) = 2x, (1,0)` Find a function `f` that has the derivative `f'(x)` and whose graph passes through the given point. Explain your reasoning.

Textbook Question

Chapter 3, 3.2 - Problem 71 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to notice that if the derivative is a linear function, then the primitive is a quadratic function, such that:

`f(x) = ax^2 + bx + c`

Differentiating f(x) yields:

`f'(x) = 2ax + b`

You need to set equal 2`x` and `2ax + b` , such that:

`2ax + b = 2x`

Comparing the coefficients of x both sides yields:

`2a = 2 => a = 1`

Notice that the free term b must be 0.

You should notice that c is not determined, yet. You may find out c using the information that the function passes through the point (1,0), such that:

`f(1) = 0 => f(1) = a*1^2 + b*1 + c `

`a + b + c = 0` 

Replacing 1 for a and 0 for b yields:

`1 + 0 + c = 0 => c = -1`

Hence, evaluating the function under the given conditions, yields `f(x) = x^2 - 1.`

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loves2learn | (Level 3) Salutatorian

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Use integration

`f(x)=int(fprime(x))dx`

`=int2xdx`

`=x^2+C`

Therefore, if the point `(1,0)` lies on the function, then `f(x)=0` when `x=1` .

From here,

You can solve for C.

`0=1^2+C`

`C=-1`

Thus, the complete function is `f(x)=x^2-1`

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