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**.

**Expert Answer**

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:

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

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