First, plot the function with with the domain restriction (on traditional graphing calculators, you can do this by changing your window settings).

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Next, identify the local (relative) extrema.

From the above graph we can see a local maximum and a local minimum (interact with the graph to see the value of the function at a certain point) . The local maximum occurs at \(x=-2.667\), and the value of the function that point is \(12.481\). The local minimum occurs at \(x=0\), and the value of the function at that point is \(3\).

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Next, evaluate the value of the function at the endpoints.

We can do this either arithmetically or by using our graphing calculator.

\[f\left(-4\right)=\left(-4\right)^{3}+4\left(-4\right)^{2}+3=-48+48+3=3\]

\[f\left(1\right)=\left(1\right)^{3}+4\left(1\right)^{2}+3=1+4+3=8\]

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Now use the above information to find the absolute (global) extrema.

The absolute maximum value of the function is \(12.481\) and it occurs at \(x=-2.667\). The absolute minimum value of the function is \(3\) and it occurs at \(x=-4\) and \(x=0\).

\[\therefore\]

1. There is a local maximum of \(12.481\) at \(x=-2.667\).
2. There is a local minimum of \(3\) at \(x=0\).
3. There is an absolute maximum of \(12.481\) at \(x=-2.667\).
4. There is an absolute minimum of \(3\) at \(x=-4\) and \(x=0\).

1. Take the derivative \[\mbox{Find the derivative using the power rule}\] \[\ f'\left(x\right)=4x^{3}-4x\]
2. Find the possible critical points \[\mbox{Factor the function and set it equal to 0}\] \[4x^{3}-4x=0\] \[4x\left(x^{2}-1\right)=0\] \[4x\left(x-1\right)\left(x+1\right)=0\] \[\therefore\] \[x=-1\mbox{, }x=0\mbox{, and }x=1\]
3. Draw a sign diagram for the derivative to determine where it is positive or negative \begin{array} {|r|r|}\hline x & x<-1 &="" x="-1" -1<x<0="" 0<x<1="">1 \\ \hline \mbox{Sign of }f'(x) &  & 0 &  & 0 &  & 0 &  \\ \hline  \end{array} To fill in the sign diagram, we will test the following specific values to see whether they are negative or positive for each interval. \[\] \[f'\left(-2\right)=4\left(-2\right)^{3}-4\left(-2\right)=-32+8=-24\] \[f'\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^{3}-4\left(-\frac{1}{2}\right)=-4\cdot\frac{1}{8}+2=1.5\] \[f'\left(\frac{1}{2}\right)=4\left(\frac{1}{2}\right)^{3}-4\left(\frac{1}{2}\right)=4\cdot\frac{1}{8}-2=-1.5\] \[f'\left(2\right)=4\left(2\right)^{3}-4\left(2\right)=32-8=24\] Now we can complete the sign diagram. \begin{array} {|r|r|}\hline x & x<-1 &="" x="-1" -1<x<0="" 0<x<1="">1 \\ \hline \mbox{Sign of }f'(x) & - & 0 & + & 0 & - & 0 & + \\ \hline  \end{array}</-1></-1>
4. Using the completed sign diagram determine where the location of the local extrema \[\]Since \(f'\left(x\right)\) is changing from negative to positive at \(x=-1\), \(x=-1\) is a local minimum. \[\]Since \(f'\left(x\right)\) is changing from positive to negative at \(x=0\), \(x=0\) is a local maximum. \[\]Since \(f'\left(x\right)\) is changing from negative to positive \(x=1\), \(x=1\) is a local minimum. \[\]
5. Find the value of the function for each local extremum. \[\] \[f\left(-1\right)=\left(-1\right)^{4}-2\left(-1\right)^{2}+2=1-2+2=1\] \[f\left(0\right)=0-0+2=2\] \[f\left(1\right)=\left(1\right)^{4}-2\left(1\right)^{2}+2=1-2+2=1\]
6. Find the absolute extrema \[\]The function is an even function (a function where it's highest power is an even number) and no domain restriction are listed; therefore, no absolute maximums will occur; however, there will be an absolute minimum, and it will be one of the local minimums. Since both local minimums are equal to each other, the absolute minimum of function occurs at both \(x=-1\mbox{ and }x=1\). \[\]
7. Summarize and list the all the extrema \[\]The function has a local maximum of \(2\) at \(x=0\). \[\]The function has a local minimum of \(1\) at \(x=-1\) and \(x=1\). \[\]The function has no absolute maximums.\[\]The function has an absolute minimum of \(1\) at \(x=-1\) and \(x=1\).

1. Find the critical points \[\mbox{To do this, set each part of the derivative equal to 0 and solve}\] \begin{array} {|r|r|}\hline x-1=0 & e^{x}-3=0 & x^{2}=0 \\ \hline  \end{array} \[x-1=0\] \[x=1\] \[e^{x}-3=0\] \[e^{x}=3\] \[\ln e^{x}=\ln\left(3\right)\] \[x=\ln\left(3\right)\] \[x^{2}=0\] \[x=0\] \[\therefore\] The derivative is equal to \(0\) when \(x=1\) and when \(x=\ln\left(3\right)\), and the derivative does not exist when \(x=0\). The previous values are possible critical points.\[\]
2. Complete a sign diagram for the critical points \begin{array} {|r|r|}\hline x & x<0 &="" x="0" 0<x<1="" 1<x<\ln{(3)}="">\ln{(3)} \\ \hline \mbox{Sign of }f' & + & \mbox{DNE} & + & 0 & - & 0 & + \\ \hline  \end{array} The specific values we tested were the following. \[h\left(-1\right)\mbox{, }h\left(\frac{1}{2}\right)\mbox{, }h\left(1.05\right)\mbox{*, and }h\left(2\right)\] You can pick any value that lies in each interval for each interval. \[\]</0>
3. Analyze the findings \[\] \(x=0\) is neither a local maximum or a local minimum since there is no sign change; therefore, \(f'\) is increasing around \(x=0\). \(x=1\) is a local maximum since the derivative changes from positive to negative. \(x=\ln{(3)}\) is a local minimum since the derivative changes from negative to positive. \[\]

Since we aren't given the original function, we are not required to find the value of the function at each extremum.

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*We had to use our calculator to find a value in between \(h\left(1\right)\mbox{ and }h\left(\ln\left(3\right)\right)\).

Instead of repeating some of the work, we will pull some answers from the last example.

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1. Find the critical points \[\] The derivative is equal to \(0\) when \(x=1\) and when \(x=\ln\left(3\right)\), and the derivative does not exist when \(x=0\). The previous values are possible critical points. \[\]
2. Find the second derivative \[\] \[f'\left(x\right)=\frac{\left(x-1\right)\left(e^{x}-3\right)}{x^{2}}\] \[f''\left(x\right)=\frac{\frac{d}{dx}\left[\left(x-1\right)\left(e^{x}-3\right)\right]\cdot x^{2}-\frac{d}{dx}\left[x^{2}\right]\cdot\left(x-1\right)\left(e^{x}-3\right)}{x^{2^{2}}}=\] \[\frac{\left[\frac{d}{dx}\left[\left(x-1\right)\right]\cdot\left(e^{x}-3\right)+\frac{d}{dx}\left[\left(e^{x}-3\right)\right]\cdot\left(x-1\right)\right]\cdot x^{2}-\left[2x\right]\cdot\left(x-1\right)\left(e^{x}-3\right)}{x^{4}}=\] \[\frac{\left[e^{x}-3+e^{x}\left(x-1\right)\right]x^{2}-2x\left(x-1\right)\left(e^{x}-3\right)}{x^{4}}\]
3. Substitute each critical point into the second derivative \[\] \[f''\left(0\right)=\frac{0-0}{0}=\mbox{undefined}\] \[\] \[f''\left(1\right)=\frac{\left[e^{1}-3+e^{1}\left(1-1\right)\right]\left(1\right)^{2}-2\left(1\right)\left(1-1\right)\left(e^{1}-3\right)}{\left(1\right)^{4}}=\] \[\frac{e-3}{1}=e-3= \mbox{some negative value}\]  since \(e\approx2.71...\), \(2.71...-3\) is a negative value; therefore, we know that the value is negative. \[\] \[f''\left(\ln\left(3\right)\right)=\frac{\left[e^{\ln\left(3\right)}-3+e^{\ln\left(3\right)}\left(\ln\left(3\right)-1\right)\right]\left(\ln\left(3\right)\right)^{2}-2\left(\ln\left(3\right)\right)\left(\ln\left(3\right)-1\right)\left(e^{\ln\left(3\right)}-3\right)}{\left(\ln\left(3\right)\right)^{4}}=\] \[\frac{\left[3-3+3\left(\ln\left(3\right)-1\right)\right]\left(\ln\left(3\right)\right)^{2}-2\left(\ln\left(3\right)\right)\left(\ln\left(3\right)-1\right)\left(3-3\right)}{\left(\ln\left(3\right)\right)^{4}}=\] \[\frac{3\left(\ln\left(3\right)-1\right)\left(\ln\left(3\right)\right)^{2}}{\left(\ln\left(3\right)\right)^{4}}=\] \[\frac{3\left(\ln\left(3\right)-1\right)}{\left(\ln\left(3\right)\right)^{2}}=\mbox{some positive number}\] From the last example we used a calculator to see that \(\ln\left(3\right)\) is great then \(1\); therefore, \(\ln\left(3\right)-1\) is a negative number, and the resulting value will be negative. We could have used a calculator to find \(f''\left(1\right)\) or \(f''\left(\ln\left(3\right)\right)\), but we didn't to show that conceptual thinking also works. \[\]
4. Analyze the findings \[\] Since \(f''\left(0\right)\) is undefined, \(f''\left(0\right)\) is neither a local maximum or a local minimum. Since \(f''\left(1\right)\) is negative, the function at \(x=1\) is concave down; therefore, a local maximum occurs at \(f\left(1\right)\). Since \(f''\left(\ln\left(3\right)\right)\) is positive, the function at \(x=\ln{(3)}\) is concave up; therefore, a local minimum occurs at \(f\left(\ln\left(3\right)\right)\)\[\]

Once again, since we aren't given the original function, we are not required to find the value of the function at each extremum.

1. Take the derivative and find the critical points \[\mbox{Find the derivative using the chain rule}\] \[f'\left(x\right)=\frac{d}{dx}\left[\ln\left(\sin\left(x^{2}\right)\right)\right]=\frac{\frac{d}{dx}\left[\sin\left(x^{2}\right)\right]}{\sin\left(x^{2}\right)}=\] \[\frac{\frac{d}{dx}\left[x^{2}\right]\cdot\cos\left(x^{2}\right)}{\sin\left(x^{2}\right)}=\frac{2x\cos\left(x^{2}\right)}{\sin\left(x^{2}\right)}\] \[\mbox{Set numerator equal to 0}\] \[2x\cos\left(x^{2}\right)=0\] \[2x=0\] \[x=0\] \[\cos\left(x^{2}\right)=0\] \[x^{2}=\cos^{-1}\left(0\right)\] \[x^{2}=\frac{n\pi}{2}\mbox{, n is any odd number}\] \[x=\pm\sqrt{\frac{n\pi}{2}}\] \[\mbox{Set the denominator equal to 0}\] \[\sin\left(x^{2}\right)=0\] \[x^{2}=\sin^{-1}\left(0\right)\] \[x^{2}=n\pi\] \[x=\pm\sqrt{n\pi}\] \[\mbox{Find the specific points in the given interval}\] The only critical point that occurs in the interval \((0,\sqrt{\pi})\) is \(x=\sqrt{\frac{\pi}{2}}\). \[\]
2. Determine which type of extrema the critical point is \[\] We can either create a sign diagram, or we can find the second derivative and test the concavity at the point. We will find the second derivative. \[f'\left(x\right)=\frac{2x\cos\left(x^{2}\right)}{\sin\left(x^{2}\right)}=2x\cot\left(x^{2}\right)\] \[f''\left(x\right)=\frac{d}{dx}\left[2x\right]\cdot\cot\left(x^{2}\right)+\frac{d}{dx}\left[\cot\left(x^{2}\right)\right]\cdot2x=\] \[2\cot\left(x^{2}\right)+2x\left(-\csc^{2}\left(x^{2}\right)\right)\] \[\mbox{substitute in \(x=\sqrt{\frac{\pi}{2}}\)}\] \[f''\left(\sqrt{\frac{\pi}{2}}\right)=\frac{2\cos\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)}-\frac{2\left(\sqrt{\frac{\pi}{2}}\right)}{\sin^{2}\left(\frac{\pi}{2}\right)}=-2\sqrt{\frac{\pi}{2}}\] Since the value we obtain is negative the function is concave down at that point; therefore, \(x=\sqrt{\frac{\pi}{2}}\) is a local maximum. \[\]
3. Find the value of the function at the critical point, and summarize your findings \[f\left(\sqrt{\frac{\pi}{2}}\right)=\ln\left(\sin\left(\sqrt{\frac{\pi}{2}}\right)^{2}\right)=\ln\left(1\right)=0\] \[\] \[\therefore\] There is a local maximum of \(0\) at \(x=\sqrt{\frac{\pi}{2}}\).