 # Find a function f such that f'(x)=3x^3 and the line 81x+y=0 is tangent to the graph of f. The aim of the question is to find the function whose first derivative is given as well as the equation tangent to it.

The basic concept behind this question is the knowledge of calculus precisely derivatives, integrals, equations of the slope, and linear equations.

The derivative of the required equation is given as:

$f^\prime\left(x\right) = 3x^3$

Given the tangent of the function, $f (x)$ is:

$81x+y=0$

As we know, the slope of the tangent can be calculated as:

$slope =\dfrac{-a}{b}$

$slope =\dfrac{-81}{1}$

$f^\prime =-81$

Putting it equal to the above equation:

$3x^3 =-81$

$x^3 =\dfrac{-81}{3}$

$x^3 =-27$

$x =-3$

Substituting the value of $x$ in the equation:

$81 x + y =0$

$81 (-23) +y=0$

$-243 + y =0$

We get the value of $y$:

$y= 243$

So, we get:

$(x,y)=(-3,243)$

Integrating the given derivative of the function:

$\int{f^\prime\left(x\right)} = \int{ 3x^3}$

$f\left(x\right) = \dfrac {3x}{4} + c$

Now to find the value of the constant $c$, let’s put the values of both the coordinates $x$ and $y$ in the above equation:

$243 =\dfrac {3(-3)}{4} + c$

$243 = \dfrac {3(81)}{4}+ c$

$243 = \dfrac {243}{4} + c$

$c = \dfrac {243}{4} -243$

$c = \dfrac {243-729}{4}$

$c = \dfrac {729}{4}$

Thus, we get the value of the constant $c$ as:

$c = \dfrac {729}{4}$

Putting it in the above equation, we get:

$f\left(x\right) = \dfrac {3x^4}{4} + c$

$f\left(x\right) = \dfrac {3x^4}{4} + \dfrac {729}{4}$

## Numerical results

Our required function is given as follows:

$f\left(x\right) = \dfrac {3x^4}{4} + \dfrac {729}{4}$

## Example

Find the function for which $f^\prime\left(x\right) = 3x^2$ and the line tangent to it is $-27x+y=0$

The derivative of the required equation is given as:

$f^\prime\left(x\right) = 3x^2$

Given the tangent of the function, $f (x)$ is:

$27x+y=0$

As we know, the slope of the tangent can be calculated as:

$slope =\dfrac {-a}{b}$

$slope =\dfrac {27}{1}$

$f^\prime =27$

Putting it equal to the above equation:

$3x^2 =27$

$x^2 =\dfrac {27}{3}$

$x^2 =9$

$x =3$

Substituting the value of $x$ in the equation:

$-27 x + y =0$

$-27 (3) +y=0$

$-81 + y =0$

We get the value of $y$:

$y= 81$

So, we get:

$(x,y)=(3, 81)$

Integrating the given derivative of the function:

$\int{f^\prime\left(x\right)} = \int{ 3x^2}$

$f\left(x\right) = \dfrac {3x^3}{3} + c$

Now to find the value of the constant $c$, let’s put the values of both the coordinates $x$ and $y$ in the above equation:

$81 = \dfrac {3\times 3^3}{3} + c$

$c = -54$

Thus, we get the value of the constant $c$ as:

$c = -54$

Putting it in the equation above, we get:

$f\left(x\right) = \dfrac {3x^3}{3} + c$

$f\left(x\right) = \dfrac {3x^3}{3} -54$