How to find (f) if f′(x)=16x^3+6x+7 and f(1)=−1 ?

3 Answers
Apr 4, 2018

f(x) = 4x^4 + 3x^2 + 7x - 15f(x)=4x4+3x2+7x15

Explanation:

We start by integrating both sides.

f(x) = int 16x^3 + 6x + 7 dxf(x)=16x3+6x+7dx

f(x) = 4x^4 + 3x^2 + 7x + Cf(x)=4x4+3x2+7x+C

Now we solve for CC.

-1 = 4(1)^4 + 3(1)^2 + 7(1) + C1=4(1)4+3(1)2+7(1)+C

-1 = 4 + 3 + 7 + C1=4+3+7+C

-1 - 14 = C114=C

C = -15C=15

Hopefully this helps!

Apr 4, 2018

f(x)=4x^4+3x^2+7x-15f(x)=4x4+3x2+7x15

Explanation:

Integrate:
16x^316x3 becomes 4x^44x4
6x6x becomes 3x^23x2
77 becomes 7x7x

So f(x)f(x) is 4x^4+3x^2+7x+C4x4+3x2+7x+C.
Plug in x=1x=1:

4(1^4)+3(1^2)+7(1)+C4(14)+3(12)+7(1)+C
=4+3+7+C=4+3+7+C
=14+C=14+C

Set 14+C14+C equal to -11:
-1=14+C1=14+C

Solve for C:
-15=C15=C

So your f(x) = 4x^4 + 3x^2+7x-15f(x)=4x4+3x2+7x15

Apr 4, 2018

f(x)=4x^4+3x^2+7x-15f(x)=4x4+3x2+7x15

Explanation:

We got:

f'(x)=16x^3+6x+7

f(1)=-1

And so,

f(x)=intf'(x) \ dx

=int16x^3+6x+7 \ dx

=4x^4+3x^2+7x+C

Therefore,

4(1)^4+3(1)^2+7*1+C=-1

4*1+3*1+7+C=-1

4+3+7+C=-1

14+C=-1

C=-15

So, the original function f(x) is:

color(blue)(f(x)=barul|4x^4+3x^2+7x-15|)