The derivative of a function $f$ is given by $f^{\prime}(x)=e^{x}-x^{3}$ . $\newline$ On which intervals is the graph of $f$ decreasing? $\newline$ Use a graphing calculator. $\newline$ Choose $1$ answer: $\newline$ (A) $[1.857,4.536]$ $\newline$ (B) $(-\infty, 1.857]$ and $[4.536, \infty)$ $\newline$ (C) $(-\infty,-0.459]$ and $[0.91,3.733]$ $\newline$ (D) $[-0.459,0.91]$ and $[3.733, \infty)$ $\newline$ (E) All real numbers

Full solution

Q. The derivative of a function

f

is given by

f^{\prime}(x)=e^{x}-x^{3}

\newline

On which intervals is the graph of

f

decreasing?

\newline

Use a graphing calculator.

\newline

Choose

1

answer:

\newline

(A)

[1.857,4.536]

\newline

(B)

(-\infty, 1.857]

and

[4.536, \infty)

\newline

(C)

(-\infty,-0.459]

and

[0.91,3.733]

\newline

(D)

[-0.459,0.91]

and

[3.733, \infty)

\newline

(E) All real numbers

Identify Decreasing Intervals: To find where the graph of $f$ is decreasing, we need to look for intervals where $f^{\prime}(x)$ is less than $0$ .
Plot $f'(x)$ : Using a graphing calculator, we plot $f'(x)=e^{x}-x^{3}$ and look for intervals where the graph is below the $x$ -axis.
Locate x-axis Intersections: The graph of $f'(x)$ crosses the x-axis at approximately $x=1.857$ and $x=4.536$ , which means the function changes from increasing to decreasing or vice versa at these points.
Analyze Interval Behavior: Between $x=1.857$ and $x=4.536$ , the graph of $f'(x)$ is above the x-axis, indicating $f$ is increasing on this interval.
Final Decreasing Intervals: Therefore, $f$ is decreasing on the intervals (- ext{\$\infty\)},$1. $857$ ]$ and [4.536, ext{\$\infty\)}).