Question

use a graphing calculator to graph the function. then use the graph to determine the number of turning points and the number and type (global or local, but not global) of any maximum or minimum values.

1. $f(x) = x(x + 1)(x + 3)$
2. $f(x) = (x + 1)^2(x - 1)(x - 2)$
3. $f(x) = -x(x - 2)^2$
4. $f(x) = -(x - 1)(x + 2)^3$

Explanation:

Step1: Analyze the function \( f(x) = x(x + 1)(x + 3) \)

First, expand the function: \( f(x)=x(x^{2}+4x + 3)=x^{3}+4x^{2}+3x \). The degree of the polynomial is 3. For a polynomial of degree \( n \), the maximum number of turning points is \( n - 1 \). So for degree 3, maximum turning points are \( 3-1 = 2 \).

Step2: Graph the function (using a graphing calculator)

When we graph \( y=x^{3}+4x^{2}+3x \), we observe the turning points. The derivative \( f^\prime(x)=3x^{2}+8x + 3 \). Setting \( f^\prime(x)=0 \), we solve \( 3x^{2}+8x + 3 = 0 \). The discriminant \( \Delta=64 - 36 = 28>0 \), so there are two real critical points, meaning two turning points.

Step3: Determine extrema

From the graph, we can see that one turning point is a local maximum and one is a local minimum (since the leading coefficient is positive, the graph goes from \( -\infty \) to \( +\infty \), so no global extrema, only local).

For \( f(x)=(x + 1)^{2}(x - 1)(x - 2) \):

Step1: Expand the function

\( (x^{2}+2x + 1)(x^{2}-3x + 2)=x^{4}-3x^{3}+2x^{2}+2x^{3}-6x^{2}+4x+x^{2}-3x + 2=x^{4}-x^{3}-3x^{2}+x + 2 \). Degree is 4, so maximum turning points \( 4 - 1=3 \).

Step2: Graph the function

Using a graphing calculator, we plot \( y = x^{4}-x^{3}-3x^{2}+x + 2 \). The derivative \( f^\prime(x)=4x^{3}-3x^{2}-6x + 1 \). By analyzing the graph, we find the number of turning points (3) and the types of extrema (local maxima and minima, no global as it's a degree 4 with positive leading coefficient, ends go to \( +\infty \)).

For \( f(x)=-x(x - 2)^{2} \):

Step1: Expand the function

\( -x(x^{2}-4x + 4)=-x^{3}+4x^{2}-4x \). Degree 3, maximum turning points \( 3 - 1 = 2 \).

Step2: Graph the function

Derivative \( f^\prime(x)=-3x^{2}+8x - 4 \). Discriminant \( \Delta = 64-48 = 16>0 \), two turning points. Analyze the graph for local extrema (leading coefficient negative, graph goes from \( +\infty \) to \( -\infty \), so local max and min).

For \( f(x)=-(x - 1)(x + 2)^{3} \):

Step1: Expand the function

\( -(x - 1)(x^{3}+6x^{2}+12x + 8)=-(x^{4}+6x^{3}+12x^{2}+8x - x^{3}-6x^{2}-12x - 8)=-x^{4}-5x^{3}-6x^{2}+4x + 8 \). Degree 4, maximum turning points \( 4 - 1 = 3 \).

Step2: Graph the function

Derivative \( f^\prime(x)=-4x^{3}-15x^{2}-12x + 4 \). Analyzing the graph (leading coefficient negative, ends go to \( -\infty \)) to find turning points and extrema.

Answer:

For \( f(x)=x(x + 1)(x + 3) \): Turning points: 2; 1 local maximum, 1 local minimum.
For \( f(x)=(x + 1)^{2}(x - 1)(x - 2) \): Turning points: 3; 2 local maxima, 1 local minimum (or vice - versa depending on graph).
For \( f(x)=-x(x - 2)^{2} \): Turning points: 2; 1 local maximum, 1 local minimum.
For \( f(x)=-(x - 1)(x + 2)^{3} \): Turning points: 3; 2 local maxima, 1 local minimum (or vice - versa depending on graph).