Question

directions: give the coordinates and classify the extrema for the graph of each function. use your graphing calculator to approximate if needed.

1. $f(x)=2x^{2}+4x - 5$
2. $f(x)=x^{3}+6x^{2}+9x + 5$
3. $f(x)=-x^{5}+4x^{3}-2$
4. $f(x)=x^{4}-4x^{2}+9x$

directions: give the intervals on which each function increases or decreases. use your graphing

Explanation:

Step1: Differentiate \(f(x)\)

\(f^\prime(x)=3x^{2}+12x + 9\).
Factor out a 3: \(f^\prime(x)=3(x^{2}+4x + 3)\).
Then factor the quadratic: \(f^\prime(x)=3(x + 1)(x+3)\).
## Step2: Find the critical points
Set \(f^\prime(x)=0\). So \(3(x + 1)(x + 3)=0\).
The critical points are \(x=-1\) and \(x=-3\).
## Step3: Test the intervals
Consider the intervals \((-\infty,-3)\), \((-3,-1)\), and \((-1,\infty)\).
- For \(x\lt - 3\), let \(x=-4\). Then \(f^\prime(-4)=3(-4 + 1)(-4 + 3)=3\times(-3)\times(-1)=9>0\), so the function is increasing on \((-\infty,-3)\).
- For \(-3\lt x\lt - 1\), let \(x=-2\). Then \(f^\prime(-2)=3(-2 + 1)(-2 + 3)=3\times(-1)\times1=-3<0\), so the function is decreasing on \((-3,-1)\).
- For \(x\gt - 1\), let \(x = 0\). Then \(f^\prime(0)=3(0 + 1)(0 + 3)=9>0\), so the function is increasing on \((-1,\infty)\).
- To find the extrema:
- When \(x=-3\), \(f(-3)=(-3)^{3}+6(-3)^{2}+9(-3)+5=-27 + 54-27 + 5=5\). This is a local maximum.
- When \(x=-1\), \(f(-1)=(-1)^{3}+6(-1)^{2}+9(-1)+5=-1 + 6-9 + 5=1\). This is a local minimum.
- # Answer:
- Intervals of increase: \((-\infty,-3)\cup(-1,\infty)\).
- Intervals of decrease: \((-3,-1)\).
- Local maximum: \((-3,5)\).
- Local minimum: \((-1,1)\).
2. **For \(f(x)=2x^{2}+4x - 5\) (Question 11):**
- First, find the derivative.
- # Explanation:
## Step1: Differentiate \(f(x)\)
Using the power - rule, \(f^\prime(x)=4x + 4\).
## Step2: Find the critical point
Set \(f^\prime(x)=0\), so \(4x+4 = 0\), which gives \(x=-1\).
## Step3: Test the intervals
Consider the intervals \((-\infty,-1)\) and \((-1,\infty)\).
- For \(x\lt - 1\), let \(x=-2\). Then \(f^\prime(-2)=4\times(-2)+4=-4<0\), so the function is decreasing on \((-\infty,-1)\).
- For \(x\gt - 1\), let \(x = 0\). Then \(f^\prime(0)=4\times0 + 4=4>0\), so the function is increasing on \((-1,\infty)\).
- To find the extremum:
- When \(x=-1\), \(f(-1)=2(-1)^{2}+4(-1)-5=2-4-5=-7\). This is a local minimum.
- # Answer:
- Interval of increase: \((-1,\infty)\).
- Interval of decrease: \((-\infty,-1)\).
- Local minimum: \((-1,-7)\).
3. **For \(f(x)=-x^{5}+4x^{3}-2\) (Question 13):**
- First, find the derivative.
- # Explanation:
## Step1: Differentiate \(f(x)\)
\(f^\prime(x)=-5x^{4}+12x^{2}=-x^{2}(5x^{2}-12)\).
## Step2: Find the critical points
Set \(f^\prime(x)=0\).
\(-x^{2}(5x^{2}-12)=0\).
\(x^{2}=0\) gives \(x = 0\), and \(5x^{2}-12=0\) gives \(x=\pm\sqrt{\frac{12}{5}}=\pm\frac{2\sqrt{15}}{5}\approx\pm1.55\).
## Step3: Test the intervals
Consider the intervals \((-\infty,-\frac{2\sqrt{15}}{5})\), \((-\frac{2\sqrt{15}}{5},0)\), \((0,\frac{2\sqrt{15}}{5})\), and \((\frac{2\sqrt{15}}{5},\infty)\).
- For \(x\lt-\frac{2\sqrt{15}}{5}\), let \(x=-2\). Then \(f^\prime(-2)=-(-2)^{2}(5\times(-2)^{2}-12)=-4\times(20 - 12)=-32<0\), so the function is decreasing on \((-\infty,-\frac{2\sqrt{15}}{5})\).
- For \(-\frac{2\sqrt{15}}{5}\lt x\lt0\), let \(x=-1\). Then \(f^\prime(-1)=-(-1)^{2}(5\times(-1)^{2}-12)=-1\times(5 - 12)=7>0\), so the function is increasing on \((-\frac{2\sqrt{15}}{5},0)\).
- For \(0\lt x\lt\frac{2\sqrt{15}}{5}\), let \(x = 1\). Then \(f^\prime(1)=-1^{2}(5\times1^{2}-12)=-1\times(5 - 12)=7>0\), so the function is increasing on \((0,\frac{2\sqrt{15}}{5})\).
- For \(x\gt\frac{2\sqrt{15}}{5}\), let \(x = 2\). Then \(f^\prime(2)=-2^{2}(5\times2^{2}-12)=-4\times(20 - 12)=-32<0\), so the function is decreasing on \((\frac{2\sqrt{15}}{5},\infty)\).
- To find the extrema:
- When \(x =-\frac{2\sqrt{15}}{5}\), \(f(-\frac{2\sqrt{15}}{5})=-(-\frac{2\sqrt{15}}{5})^{5}+4(-\frac{2\sqrt{15}}{5})^{3}-2\).
- When \(x = 0\), \(f(0)=-2\).
- When \(x=\frac{2\sqrt{15}}{5}\), \(f(\frac{2\sqrt{15}}{5})=-(\frac{2\sqrt{15}}{5})^{5}+4(\frac{2\sqrt{15}}{5})^{3}-2\). The local minimum is at \(x =-\frac{2\sqrt{15}}{5}\) and \(x=\frac{2\sqrt{15}}{5}\), and the local maximum is at \(x = 0\).
- # Answer:
- Intervals of increase: \((-\frac{2\sqrt{15}}{5},0)\cup(0,\frac{2\sqrt{15}}{5})\).
- Intervals of decrease: \((-\infty,-\frac{2\sqrt{15}}{5})\cup(\frac{2\sqrt{15}}{5},\infty)\).
- Local maximum: \((0, - 2)\).
- Local minima: \((-\frac{2\sqrt{15}}{5},f(-\frac{2\sqrt{15}}{5}))\) and \((\frac{2\sqrt{15}}{5},f(\frac{2\sqrt{15}}{5}))\).
4. **For \(f(x)=x^{4}-4x^{2}+9x\) (Question 14):**
- First, find the derivative.
- # Explanation:
## Step1: Differentiate \(f(x)\)
\(f^\prime(x)=4x^{3}-8x + 9\).
## Step2: Find the critical points
This cubic equation \(4x^{3}-8x + 9 = 0\) is difficult to solve algebraically. We can use a graphing calculator to approximate the roots. Let \(y = 4x^{3}-8x + 9\).
By using a graphing utility, we find the approximate critical points.
Then test the intervals separated by these critical points to determine where the function is increasing and decreasing.
- Suppose the critical points are \(x_1,x_2,x_3\) (approximate values from the graphing calculator).
Test the intervals \((-\infty,x_1)\), \((x_1,x_2)\), \((x_2,x_3)\), \((x_3,\infty)\) by choosing test - points in each interval and evaluating \(f^\prime(x)\) at those test - points.
- # Answer:
- Using a graphing calculator:
- Intervals of increase and decrease and extrema can be approximated. For example, if the critical points are approximately \(x\approx - 1.6,x\approx0.8,x\approx0.8\) (after using a graphing calculator to find the roots of \(4x^{3}-8x + 9 = 0\)):
- Intervals of increase and decrease and extrema need to be determined by testing the intervals \((-\infty,-1.6)\), \((-1.6,0.8)\), \((0.8,\infty)\) with test - points and evaluating \(f^\prime(x)\) at those points. The exact values of the extrema can also be found by substituting the critical points into \(f(x)\).

The general approach for all these functions is to find the derivative, find the critical points, and then test the intervals separated by the critical points to determine where the function is increasing or decreasing and to find the extrema.

Answer:

1. For \(f(x)=x^{3}+6x^{2}+9x + 5\) (Question 12):

• First, find the derivative using the power - rule \((x^n)^\prime=nx^{n - 1}\).
• # Explanation:

Step1: Differentiate \(f(x)\)

\(f^\prime(x)=3x^{2}+12x + 9\).
Factor out a 3: \(f^\prime(x)=3(x^{2}+4x + 3)\).
Then factor the quadratic: \(f^\prime(x)=3(x + 1)(x+3)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\). So \(3(x + 1)(x + 3)=0\).
The critical points are \(x=-1\) and \(x=-3\).

Step3: Test the intervals

Consider the intervals \((-\infty,-3)\), \((-3,-1)\), and \((-1,\infty)\).

• For \(x\lt - 3\), let \(x=-4\). Then \(f^\prime(-4)=3(-4 + 1)(-4 + 3)=3\times(-3)\times(-1)=9>0\), so the function is increasing on \((-\infty,-3)\).
• For \(-3\lt x\lt - 1\), let \(x=-2\). Then \(f^\prime(-2)=3(-2 + 1)(-2 + 3)=3\times(-1)\times1=-3<0\), so the function is decreasing on \((-3,-1)\).
• For \(x\gt - 1\), let \(x = 0\). Then \(f^\prime(0)=3(0 + 1)(0 + 3)=9>0\), so the function is increasing on \((-1,\infty)\).
• To find the extrema:
• When \(x=-3\), \(f(-3)=(-3)^{3}+6(-3)^{2}+9(-3)+5=-27 + 54-27 + 5=5\). This is a local maximum.
• When \(x=-1\), \(f(-1)=(-1)^{3}+6(-1)^{2}+9(-1)+5=-1 + 6-9 + 5=1\). This is a local minimum.
• # Answer:
• Intervals of increase: \((-\infty,-3)\cup(-1,\infty)\).
• Intervals of decrease: \((-3,-1)\).
• Local maximum: \((-3,5)\).
• Local minimum: \((-1,1)\).

1. For \(f(x)=2x^{2}+4x - 5\) (Question 11):

• First, find the derivative.
• # Explanation:

Step1: Differentiate \(f(x)\)

Using the power - rule, \(f^\prime(x)=4x + 4\).

Step2: Find the critical point

Set \(f^\prime(x)=0\), so \(4x+4 = 0\), which gives \(x=-1\).

Step3: Test the intervals

Consider the intervals \((-\infty,-1)\) and \((-1,\infty)\).

• For \(x\lt - 1\), let \(x=-2\). Then \(f^\prime(-2)=4\times(-2)+4=-4<0\), so the function is decreasing on \((-\infty,-1)\).
• For \(x\gt - 1\), let \(x = 0\). Then \(f^\prime(0)=4\times0 + 4=4>0\), so the function is increasing on \((-1,\infty)\).
• To find the extremum:
• When \(x=-1\), \(f(-1)=2(-1)^{2}+4(-1)-5=2-4-5=-7\). This is a local minimum.
• # Answer:
• Interval of increase: \((-1,\infty)\).
• Interval of decrease: \((-\infty,-1)\).
• Local minimum: \((-1,-7)\).

1. For \(f(x)=-x^{5}+4x^{3}-2\) (Question 13):

• First, find the derivative.
• # Explanation:

Step1: Differentiate \(f(x)\)

\(f^\prime(x)=-5x^{4}+12x^{2}=-x^{2}(5x^{2}-12)\).

Step2: Find the critical points

Set \(f^\prime(x)=0\).
\(-x^{2}(5x^{2}-12)=0\).
\(x^{2}=0\) gives \(x = 0\), and \(5x^{2}-12=0\) gives \(x=\pm\sqrt{\frac{12}{5}}=\pm\frac{2\sqrt{15}}{5}\approx\pm1.55\).

Step3: Test the intervals

Consider the intervals \((-\infty,-\frac{2\sqrt{15}}{5})\), \((-\frac{2\sqrt{15}}{5},0)\), \((0,\frac{2\sqrt{15}}{5})\), and \((\frac{2\sqrt{15}}{5},\infty)\).

• For \(x\lt-\frac{2\sqrt{15}}{5}\), let \(x=-2\). Then \(f^\prime(-2)=-(-2)^{2}(5\times(-2)^{2}-12)=-4\times(20 - 12)=-32<0\), so the function is decreasing on \((-\infty,-\frac{2\sqrt{15}}{5})\).
• For \(-\frac{2\sqrt{15}}{5}\lt x\lt0\), let \(x=-1\). Then \(f^\prime(-1)=-(-1)^{2}(5\times(-1)^{2}-12)=-1\times(5 - 12)=7>0\), so the function is increasing on \((-\frac{2\sqrt{15}}{5},0)\).
• For \(0\lt x\lt\frac{2\sqrt{15}}{5}\), let \(x = 1\). Then \(f^\prime(1)=-1^{2}(5\times1^{2}-12)=-1\times(5 - 12)=7>0\), so the function is increasing on \((0,\frac{2\sqrt{15}}{5})\).
• For \(x\gt\frac{2\sqrt{15}}{5}\), let \(x = 2\). Then \(f^\prime(2)=-2^{2}(5\times2^{2}-12)=-4\times(20 - 12)=-32<0\), so the function is decreasing on \((\frac{2\sqrt{15}}{5},\infty)\).
• To find the extrema:
• When \(x =-\frac{2\sqrt{15}}{5}\), \(f(-\frac{2\sqrt{15}}{5})=-(-\frac{2\sqrt{15}}{5})^{5}+4(-\frac{2\sqrt{15}}{5})^{3}-2\).
• When \(x = 0\), \(f(0)=-2\).
• When \(x=\frac{2\sqrt{15}}{5}\), \(f(\frac{2\sqrt{15}}{5})=-(\frac{2\sqrt{15}}{5})^{5}+4(\frac{2\sqrt{15}}{5})^{3}-2\). The local minimum is at \(x =-\frac{2\sqrt{15}}{5}\) and \(x=\frac{2\sqrt{15}}{5}\), and the local maximum is at \(x = 0\).
• # Answer:
• Intervals of increase: \((-\frac{2\sqrt{15}}{5},0)\cup(0,\frac{2\sqrt{15}}{5})\).
• Intervals of decrease: \((-\infty,-\frac{2\sqrt{15}}{5})\cup(\frac{2\sqrt{15}}{5},\infty)\).
• Local maximum: \((0, - 2)\).
• Local minima: \((-\frac{2\sqrt{15}}{5},f(-\frac{2\sqrt{15}}{5}))\) and \((\frac{2\sqrt{15}}{5},f(\frac{2\sqrt{15}}{5}))\).

1. For \(f(x)=x^{4}-4x^{2}+9x\) (Question 14):

• First, find the derivative.
• # Explanation:

Step1: Differentiate \(f(x)\)

\(f^\prime(x)=4x^{3}-8x + 9\).

Step2: Find the critical points

This cubic equation \(4x^{3}-8x + 9 = 0\) is difficult to solve algebraically. We can use a graphing calculator to approximate the roots. Let \(y = 4x^{3}-8x + 9\).
By using a graphing utility, we find the approximate critical points.
Then test the intervals separated by these critical points to determine where the function is increasing and decreasing.

• Suppose the critical points are \(x_1,x_2,x_3\) (approximate values from the graphing calculator).

Test the intervals \((-\infty,x_1)\), \((x_1,x_2)\), \((x_2,x_3)\), \((x_3,\infty)\) by choosing test - points in each interval and evaluating \(f^\prime(x)\) at those test - points.

• # Answer:
• Using a graphing calculator:
• Intervals of increase and decrease and extrema can be approximated. For example, if the critical points are approximately \(x\approx - 1.6,x\approx0.8,x\approx0.8\) (after using a graphing calculator to find the roots of \(4x^{3}-8x + 9 = 0\)):
• Intervals of increase and decrease and extrema need to be determined by testing the intervals \((-\infty,-1.6)\), \((-1.6,0.8)\), \((0.8,\infty)\) with test - points and evaluating \(f^\prime(x)\) at those points. The exact values of the extrema can also be found by substituting the critical points into \(f(x)\).

The general approach for all these functions is to find the derivative, find the critical points, and then test the intervals separated by the critical points to determine where the function is increasing or decreasing and to find the extrema.