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

Apr 4, 2018

$f \left(x\right) = 4 {x}^{4} + 3 {x}^{2} + 7 x - 15$

#### Explanation:

We start by integrating both sides.

$f \left(x\right) = \int 16 {x}^{3} + 6 x + 7 \mathrm{dx}$

$f \left(x\right) = 4 {x}^{4} + 3 {x}^{2} + 7 x + C$

Now we solve for $C$.

$- 1 = 4 {\left(1\right)}^{4} + 3 {\left(1\right)}^{2} + 7 \left(1\right) + C$

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

$- 1 - 14 = C$

$C = - 15$

Hopefully this helps!

Apr 4, 2018

$f \left(x\right) = 4 {x}^{4} + 3 {x}^{2} + 7 x - 15$

#### Explanation:

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

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

$4 \left({1}^{4}\right) + 3 \left({1}^{2}\right) + 7 \left(1\right) + C$
$= 4 + 3 + 7 + C$
$= 14 + C$

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

Solve for C:
$- 15 = C$

So your $f \left(x\right) = 4 {x}^{4} + 3 {x}^{2} + 7 x - 15$

Apr 4, 2018

$f \left(x\right) = 4 {x}^{4} + 3 {x}^{2} + 7 x - 15$

#### Explanation:

We got:

$f ' \left(x\right) = 16 {x}^{3} + 6 x + 7$

$f \left(1\right) = - 1$

And so,

$f \left(x\right) = \int f ' \left(x\right) \setminus \mathrm{dx}$

$= \int 16 {x}^{3} + 6 x + 7 \setminus \mathrm{dx}$

$= 4 {x}^{4} + 3 {x}^{2} + 7 x + C$

Therefore,

$4 {\left(1\right)}^{4} + 3 {\left(1\right)}^{2} + 7 \cdot 1 + C = - 1$

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

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

$14 + C = - 1$

$C = - 15$

So, the original function $f \left(x\right)$ is:

$\textcolor{b l u e}{f \left(x\right) = \overline{\underline{|}} 4 {x}^{4} + 3 {x}^{2} + 7 x - 15 |}$