`f'(x) = 6x - 1, (2,7)` 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 72 - 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 6x-1 and 2ax + b, such that:

`2ax + b = 6x-1`

Comparing the coefficients of x both sides yields:

`2a =6 => a = 3`

Notice that the free term b must be -1.

You should notice that c is not determined, yet. You may find out c using the information that the function passes through the point (2,7), such that:

`f(2) =7 => f(2) = a*2^2 + b*2 + c`

`4a + 2b + c = 7`

Replacing 3 for a and -1 for b yields:

`12- 2 + c =7 => c = 7-10 => c = -3`

Hence, evaluating the function under the given conditions, yields `f(x) = 3x^2 - x - 3.`

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 2x 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)=intfprime(x)dx`

`=int6x-1dx`

`=3x^2-x+C`

Therefore, if the point `(2,7)` lies on the function, then `f(x)=7` when `x=2` .

From here,

You can solve for C.

`7=3(2)^2-2+C`

`C=-3`

Thus, the complete function is `f(x)=3x^2-x-3`

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