Question

Local Maximum and Minimum Values A function is given. (a) Find all the local maximum and minimum values of the function and the value of x at which each occurs. State each answer rounded to two decimal places. (b) Find the intervals on which the function is increasing and on which the function is decreasing. State each answer rounded to two decimal places.

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

Short answer

(a) The local maximum value is$4$ at $x=1$.The local maximum value is$2.815$ at $x=-0.333$.

(b) The function is $f$increasing on $\left(\frac{-1}{3},1\right)$.

The function is $f$decreasing on $\left(-\infty ,\frac{-1}{3}\right)U\left(1,\infty \right)$.

Step-by-step solution

Part a. Step 1. Plotting the graph.

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

Part a. Step 2. Concept of local maximum and local minimum.

The function value $f\left(a\right)$is a local maximum value of $f$if $f\left(a\right)\ge f\left(x\right)$when $x$ is a local minimum value of if$f\left(a\right)\le f\left(x\right)$ when$x$ is a near $a$.

Part a. Step 3. Determining the local maximum and local minimum value.

The local maximum value is$4$ at $x=1$. The local maximum value is$2.815$ at $x=-0.333$.

Part b. Step 1. Plotting the graph.

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

Part b. Step 2. Concept of increasing and decreasing of function.

The function$f$ is said to be increasing when the value of$y$ continuously increases that is $f\left(y_1\right)<f\left(y_2\right)$.

Where $y_1$and$y_2$ is at $y$–axis such that$y_1<y_2$.

The function$f$ is said to be decreasing when the value of$y$ continuously decreases that is $f\left(y_1\right)>f\left(y_2\right)$.

Where $y_1$and$y_2$ is at $y$–axis such that$y_1<y_2$.

Part b. Step 3. Determining the interval on which f is increasing and decreasing.

The function is $f$increasing on $\left(\frac{-1}{3},1\right)$.

The function is $f$decreasing on $\left(-\infty ,\frac{-1}{3}\right)U\left(1,\infty \right)$.