Question

Increasing and Decreasing A function f is given. (a) Use a graphing calculator to draw the graph of f. (b) Find the domain and range of f. (c) State approximately the intervals on which f is increasing and on which f is decreasing.

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

Short answer

The domain of f is$IR$and range of f is$IR$.

The function of f is increasing on $(-\infty ,\frac{-2+\sqrt{7}}{3})\cup (\frac{-2-\sqrt{7}}{3},\infty )$.

The function of f is decreasing on $(-\frac{2+\sqrt{7}}{3},-\frac{2-\sqrt{7}}{3})$.

Step-by-step solution

Part a. Step 1. Concept of plotting the graph.

Here, we plot the graph in which abscissa is the $x$-axis and ordinate is $y$-axis.

Part a. Step 2. Concept of the coordinate axis.

For plotting graph we move right for positive $x$-axis and left for negative $x$-axis and we move upward for positive $y$-axis and downward for negative $y$-axis.

Part a. Step 3. Plotting the graph.

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

Part b. Step 1. Concept of domain.

The domain of a function is the set of values that we are allowed to plug into our function.

Part b. Step 2. Concept of range.

The range of a function is the set of values that the function assumes. This set is the values that the function shoots out after we plug an$x$ value in.

Part b. Step 3. Determining the domain and range of a function.

Domain of$f\left(x\right)=x^3+2x^2-x-2$s$IR$.

Range of $f\left(x\right)=x^3+2x^2-x-2$s $IR$.

Part c. Step 1. Given function.

Here, the given function,$f\left(x\right)=x^3+2x^2-x-2$

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

The function of f is increasing on $(-\infty ,\frac{-2+\sqrt{7}}{3})\cup (\frac{-2-\sqrt{7}}{3},\infty )$and the function of f is decreasing on $(-\frac{2+\sqrt{7}}{3},-\frac{2-\sqrt{7}}{3})$.