Chapter 4.9, Problem 1PT

To determine

Whether the statement “If $h^{\prime }\left(x\right)=k\left(x\right)$, then $k\left(x\right)$ is an antiderivative of $h\left(x\right)$” is true or false.

Answer to Problem 1PT

The given statement is $\underset{\_}{\text{false}}$.

Explanation of Solution

The given statement is, “If $h^{\prime }\left(x\right)=k\left(x\right)$, then $k\left(x\right)$ is an antiderivative of $h\left(x\right)$”.

Equivalently if $h^{\prime }\left(x\right)=k\left(x\right)$, then $k^{\prime }\left(x\right)=h\left(x\right)$.

The following example disproves the given statement.

Consider the derivative function $h^{\prime }\left(x\right)=4x^3$.

On comparing this with the given statement, it is obtained that, the function is $k\left(x\right)=4x^3$.

Compute the value of $k^{\prime }\left(x\right)$ as follows.

$\begin{array}{c}{k}^{\prime }\left(x\right)=\frac{d}{dx}\left(4x^3\right)\\ =4\left(3x^2\right)\\ =12x^2\end{array}$

Thus, the value of $k^{\prime }\left(x\right)$ is, $12x^2$.

Now find the value of $h\left(x\right)$ as follows.

$\begin{array}{c}h\left(x\right)={\displaystyle \int h^{\prime }\left(x\right)dx}\\ ={\displaystyle \int 4x^{3}dx}\\ =\frac{4x^4}{4}+C\\ =x^4+C\end{array}$

Thus, the value of $h\left(x\right)$ is $x^4+C$.

Clearly, it is observed that $k^{\prime }\left(x\right)\ne h\left(x\right)$.

Therefore, the given statement is $\underset{\_}{\text{false}}$.