# Find f(x) if f'(x) = 2x^2 - 5x + 7 and f(0) = -3

Asked on by lalooo

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have f'(x) = 2x^2 - 5x + 7   and f(0) = -3.

Now let's integrate f'(x)

=> Int [ 2x^2 - 5x + 7 ]

=> 2x^3/3 - 5x^2/2 + 7x + C

Now f(0) = -3

=> 2x^3/3 - 5x^2/2 + 7x + C = -3

=> C = -3

So f(x) = 2x^3/3 - 5x^2/2 + 7x -3

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

Find f(x) if f'(x) = 2x^2 - 5x + 7   and f(0) = -3

We know that:

f(x) = integral f'(x)

==> f(x) = intg ( 2x^2 - 5x + 7) dx

= intg (2x^2) dx - intg (5x) dx + intg 7 dx

= 2x^3/3  - 5x^2 /2 + 7x + C

= (2/3)x^3 - (5/2)x^2 + 7x + C

But given f(0) = -3

Then :

We will substitute with x= 0 :

f(0) = (2/3)*0 - (5/2)*0 + 7*0 + C = -3

==> C = -3

==> f(x) = (2/3)x^3 - (5/2)x^2 + 7x - 3

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Given f'(x) = 2x^2-5x-7 to find f(x), if  f(0) = -3.

First we find the f(x).

Since f'(x) is given, f(x) = Int f'(x) dx.

Therefore f(x) = Int (2x^2-5x+7) dx.

f(x) = Int 2x^2dx - Int5xdx + Int 7 dx.

f(x) =  2 (1/3)x^3 - 5(1/2)x^2 +7x +Const , as Int Kx^n dx = K(1/n+1)x^(n+1) , for all n except n = -1.

Therefore f(x) = (2/3)x^3-(5/2)x^2+7x +C....(1)

Now we use the condition f(0) = -3 to find C:

f(0) = (2/3)*0^3-(5/2)*0^2+7*0 +C = -3.

Therefore C = -3.

So we rewrite  f(x) in (1) replacing C by -3:

f(x) = (2/3)x^3-(5/2)x^2+7x -3.

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