1. Take the derivative using the power rule \[f'\left(x\right)=4x+4\]
2. Set the derivative equal to \(0\) and solve for \(x\) \[4x+4=0\] \[4x=-4\] \[\mbox{Divide both sides by 4 and solve}\] \[x=-1\]
3. Determine where the derivative does not exist \[\] Since the derivative is a straight line, it doesn't consist of any discontinuities; therefore, the derivative exists everywhere.\[\]
4. Determine whether the original function exists at the possible critical points \[\] Since the \(f'\left(1\right)\) exists, the original function will also exist at \(x=1\).\[\]
5. Summarize your findings \[\] The critical point occurs at \(x=-1\).

1. Take the derivative using the power rule \[f'\left(x\right)=6x^{2}+30x+36\]
2. Set the derivative equal to \(0\) and solve for \(x\) \[6x^{2}+30x+36=0\] \[\mbox{Factor out 6}\] \[6\left(x^{2}+5x+6\right)=0\] \[\mbox{Divide both sides by 6}\] \[x^{2}+5x+6=0\] \[\mbox{Factor and solve}\] \[\left(x+2\right)\left(x+3\right)=0\] \[x=-2,\: x=-3\]
3. Determine where the derivative does not exist \[\] Since the derivative is a parabola, it doesn't consist of any discontinuities; therefore, the derivative exists everywhere. \[\]
4. Determine whether the original function exists at the possible critical points \[\] Since \(f'\left(-2\right)\) and \(f'\left(3\right)\)-2\) exist, the original function will also exist at both \(x=-2\) and \(x=3\). \[\]
5. Summarize your findings \[\] The critical points occur at \(x=-2\) and \(x=-3\).

1. Take the derivative using the power rule \[f'\left(x\right)=x^{3}+3x^{2}-4x^{2}-12\]  
2. Set the derivative equal to \(0\) and solve for \(x\) \[x^{3}+3x^{2}-4x-12=0\] \[\mbox{Factor out \(x^{2}\) from the first two terms and -4 from the last two terms}\] \[x^{2}\left(x+3\right)-4\left(x+3\right)=0\] \[\mbox{Factor out the term (x+3)}\] \[\left(x+3\right)\left(x^{2}-4\right)=0\] \[\mbox{Factor the term \((x^{2}-4)\) and solve}\] \[\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\] \[x=-3,\:x=-2, \:x=2\]
3. Determine where the derivative does not exist \[\] Since the derivative is a cubic function, it doesn't consist of any discontinuities; therefore, the derivative exists everywhere.\[\]
4. Determine whether the original function exists at the possible critical points \[\] Since \(f'(-3)\), \(f'(-2)\) and \(f'(2)\) exist, the original function will also exist at \(x=-3\), \(x=-2\), and \(x=2\). \[\]
5. Summarize your findings \[\] The critical points occur at \(x=-3\), \(x=-2\), and \(x=2\).

First, take the derivative using the quotient rule.

\[f'\left(x\right)=\frac{\left(x-3\right)\frac{d}{dx}\left[2x^{2}\right]-\left(2x^{2}\right)\frac{d}{dx}\left[x-3\right]}{\left(x-3\right)^{2}}=\] \[\frac{\left(x-3\right)\left(4x\right)-\left(2x^{2}\right)\left(1\right)}{\left(x-3\right)^{2}}\]

Since the derivative function is a rational function, it will be equal to \(0\) when the numerator is equal to \(0\), and it will be discontinuous when the denominator is equal to \(0\); therefore, we will first simply the above expression and separately set both the numerator and the denominator equal to \(0\).

1. Find where the derivative function equals \(0\) \[\mbox{Expand the numerator}\] \[f'\left(x\right)=\frac{\left(x-3\right)\left(4x\right)-\left(2x^{2}\right)\left(1\right)}{\left(x-3\right)^{2}}=\frac{4x^{2}-12x-2x^{2}}{\left(x-3\right)^{2}}=\]\[\mbox{Simplify the numerator}\] \[\frac{2x^{2}-3x}{\left(x-3\right)^{2}}=\frac{2x\left(x-6\right)}{\left(x-3\right)^{2}}\] \[\mbox{Set the numerator equal to 0 and solve}\] \[2x\left(x-6\right)=0\] \[x=0,\:x=6\]\[\mbox{Set the numerator equal to 0 and solve}\]
2. Find where the derivative function does not exist \[\mbox{Set the denominator equal to 0 and solve}\] \[\left(x-3\right)^{2}=0\] \[x=3\]
3. Determine whether the original function exists at the possible critical points \[\] Since \(f'(0)\) and \(f'(6)\) exist, the original function will also exist at \(x=0\) and \(x=6\). On the other note, the original function is discontinuous at \(x=3\), because at \(x=3\) the denominator of the original function is equal to \(0\); therefore, \(x=3\) is not a critical point. \[\]
4. Summarize your findings \[\] The critical points occur at \(x=0\) and \(x=6\).

1. Take the derivative using the product and chain rule \[f'\left(x\right)=\frac{d}{dx}\left[x^{2}\right]e^{-3x}+\frac{d}{dx}\left[e^{-3x}\right]x^{2}=\] \[2xe^{-3x}+\left(-3\right)e^{-3x}\cdot x^{2}=2xe^{-3x}-3x^{2}e^{-3x}\]
2. Set the derivative equal to \(0\) and solve for \(x\) \[\] The exponential function can never be equal to \(0\); therefore, we can factor it out of the expression and divide both sides by it. \[2xe^{-3x}-3x^{2}e^{-3x}=0\] \[e^{-3x}\left[2x-3x^{2}\right]=0\] \[2x-3x^{2}=0\] Now we can factor the above expression and solve for \(x\) \[x\left(2-3x\right)=0\] \[x=0,\: x=\frac{2}{3}\]
3. Determine where the derivative does not exist \[\] If we look at the following expression, we can see that there are two functions that make up the product. The first function is an exponential function, which is always continuous, and the second function is a polynomial, which is also always continuous; therefore, the two functions multiplied together will also be continuous everywhere, which implies that the derivative function is never discontinuous. \[e^{-3x}\left(2x-3x^{2}\right)=0\]
4. Determine whether the original function exists at the possible critical points \[\] Since \(f'(0)\) and \(f'(\frac{2}{3})\) exist, the original function will also exist at both \(x=0\) and \(x=\frac{2}{3}\). \[\]
5. Summarize your findings \[\] The critical points occur at \(x=0\) and \(x=\frac{2}{3}\).

1. Find the derivative \[\] Before we take the derivative of an absolute value function, we need to write in the piecewise function format. \[f(x) = \left\{\begin{aligned} &3x-2, && x\geq2\\ &-3x+2, && x
2. Find where the derivative is equal to \(0\) \[\] The derivative is a piecewise function that is made of two horizontal lines; therefore, the derivative will never be equal to \(0\) \[y=\pm3 \neq 0\]
3. Determine where the derivative does not exist \[\] Both parts of the piecewise function are horizontal lines and separately they are always continuous; however, in the piecewise function they are joined at \(x=2\) and since \(\mathop {\lim }\limits_{x \to 2^+} f'\left( x \right)\neq\mathop {\lim }\limits_{x \to 2^-} f'\left( x \right)\), the derivative function is discontinuous at \(x=2\).\[\]
4. Determine whether the original function exists at the possible critical points \[\] The derivative function does not exist at \(x=2\); however, since the original function is continuous everywhere, it exists at \(x=2\).[\]
5. Summarize your findings \[\] The only critical point occurs at \(x=2\).

1. Take the derivative using the trigonometric derivative rules \[\] \[f'\left(x\right)=\sec^{2}\left(x\right)+\cos\left(x\right)\]
2. Compute where the derivative is equal to \(0\) on the given interval \[\] \[\sec^{2}\left(x\right)+\cos\left(x\right)=0\] \[\mbox{Rearrange the equation using trigonometric identities}\] \[\sec^{2}\left(x\right)=-\cos\left(x\right)\] \[\frac{1}{\cos^{2}\left(x\right)}=-\cos\left(x\right)\] \[1=-\cos^{3}\left(x\right)\] \[\sqrt[3]{1}=\sqrt[3]{-\cos^{3}\left(x\right)}\] \[1=-\cos\left(x\right)\] \[\cos\left(x\right)=-1\] We we con solve for \(x\) \[x=\cos^{-1}\left(-1\right)\] The cosine function is equal to \(-1\) at \(x=n\pi\) where \(n\) is an odd number; therefore, on the given interval, the cosine function will be equal to \(-1\) at \(x=\pi\) and \(x=-\pi\). Moreover, the derivative function is equal to \(0\) at \(x=\pi\) and \(x=-\pi\)\[\]
3. Determine where the derivative does not exist \[\]Of the two function that make up the derivative function, the cosine function is always be continuous; therefore, we can ignore it and focus on the secant squared function. The secant function is the reciprocal of the cosine function; therefore, we need to determine where cosine is equal to \(0\) on the given interval. Cosine is equal to \(0\) at \(x=\frac{n\pi}{2}\) where \(n\) is any odd number. That means that in our given interval cosine is equal to \(0\) at \(x=\frac{\pi}{2}\) and \(x=-\frac{\pi}{2}\) and at those points the secant function will be discontinuous; therefore, the derivative function is discontinuous at \(x=\frac{\pi}{2}\) and \(x=-\frac{\pi}{2}\).\[\]
4. Determine whether the original function exists at the possible critical points \[\] Since the tangent function is discontinuous at \(x=\pm\frac{n\pi}{2}\) (where n is an odd number), the original function is discontinuous at \(x=\frac{\pi}{2}\) and \(x=-\frac{\pi}{2}\); therefore, \(x=\frac{\pi}{2}\) and \(x=-\frac{\pi}{2}\) are not critical points.\[\]
5. Summarize your findings \[\] The critical points occur at \(x=-\pi\) and \(x=\pi\).