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

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Answer 1 · 2018-04-04T03:44:38

$f(x) = 4x^4 + 3x^2 + 7x - 15$

We start by integrating both sides.

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

$f(x) = 4x^4 + 3x^2 + 7x + C$

Now we solve for $C$ .

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

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

$-1 - 14 = C$

$C = -15$

Hopefully this helps!

Answer 2 · 2018-04-04T03:51:50

$f(x)=4x^4+3x^2+7x-15$

Integrate:
$16x^3$ becomes $4x^4$
$6x$ becomes $3x^2$
$7$ becomes $7x$

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

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

Set $14+C$ equal to $-1$ :
$-1=14+C$

Solve for C:
$-15=C$

So your $f(x) = 4x^4 + 3x^2+7x-15$

Answer 3 · 2018-04-04T08:32:33

$f(x)=4x^4+3x^2+7x-15$

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:

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