The \(y\)-intercept is \((0,\answer [given]{0})\). Place this point on your plot.

Which of the following are vertical asymptotes? Select all that apply.

In this case, \(f(x) =2x^3-3x^2-12x\), we can find the \(x\)-intercepts. There are three \(x\) intercepts. Call them \((a,0)\), \((b,0)\), and \((c,0)\), and order them such that \(a<b<c\). Then

\begin{align*} a &= \answer [given]{\frac {3-\sqrt {105}}{4}},\\ b &= \answer [given]{0}, \\ c &= \answer [given]{\frac {3+\sqrt {105}}{4}}. \end{align*}

Which of the following best describes the end behavior of \(f\) as \(x \to \infty \)?

Compute \(f'(x)\) and \(f''(x)\), \begin{align*} f'(x) &= \answer [given]{6x^2 -6x -12}\\ &=6\left (\answer [given]{x^2-x-2}\right )\\ &=6(x+1)\left (\answer [given]{x-2}\right ). \end{align*}

On the other hand \begin{align*} f''(x) &= \answer [given]{12x-6}\\ &=12\left (\answer [given]{x-\frac {1}{2}}\right ). \end{align*}

The critical numbers are where \(f'(x) = 0\), thus we need to solve \(6(x+1)(x-2) = 0\) for \(x\). This equation has two solutions. If we call them \(a\) and \(b\), with \(a<b\), then what are \(a\) and \(b\)?

Mark the critical numbers \(x=2\) and \(x=-1\) on your plot.

Similarly, since for \(-1<x<2\), \((x+1)>0\) and \((x-2)<0\), the derivative is negative there, and, therefore, our function is decreasing on \((-1,2)\).

And, for \(x>2\), both factors \((x+1)\) and \((x-2)\) are positive, and, therefore, our function is increasing on \((2,\infty )\).

Hence \(x=-1\), corresponding to the point \((-1,7)\), is a local maximum and \(x=2\), corresponding to the point \((2,-20)\), is local minimum of \(f(x)\). Identify this on your plot.

In order to locate inflection points, we have to solve the equation \(f''(x) = 0\), thus we need to solve \(12\left (x-\frac {1}{2}\right )=0\) for \(x\).

The solution to this is \(x = \answer {\frac {1}{2}}\).

This is only a possible inflection point; we still have to check whether concavity changes there.

We have that \(f''(x)<0\) for \(x<\frac {1}{2}\), therefore \(f\) is concave downon \(\left (-\infty ,\frac {1}{2}\right )\).

Similarly, \(f''(x)>0\) for \(x>\frac {1}{2}\), therefore \(f\) is concave up on \(\left (\frac {1}{2},\infty \right )\).

So, concavity changes at \(x=\frac {1}{2}\), and therefore, this point is a point of inflection.

Since all of this behavior as described above occurs on the interval \([-2,4]\), we now have a complete sketch of \(f(x)\) on this interval, see the figure below.

Since this function is piecewise defined, we will analyze the the graph of \(f\) on intervals \((-\infty ,0)\) and \([0,\infty )\) separately.

Because \(f\) is piecewise defined, and potentially discontinuous at \(0\), it is important to understand the behavior of \(f\) near \(x = 0\).

Moreover,

Record this information on our graph with filled and unfilled circles.

Which of the following are vertical asymptotes on \((-\infty ,0)\)? Select all that apply.

Which of the following are vertical asymptotes on \((0,\infty )\)? Select all that apply.

Which of the following best describes the end behavior of \(f\) as \(x \to \infty \)?

Which of the following best describes the end behavior of \(f\) as \(x \to -\infty \)?

We mark the location of the horizontal asymptote:

The derivative of \(f\) on \((-\infty , 0)\) is

The derivative of \(f\) on \((0, \infty )\) is

The critical numbers are the numbers in the domain where \(f'(x) = 0\) or \(f'(x)\) does not exist. \(0\) is a critical number, since we have already seen it is a discontinuity for \(f\), and thus \(f'(0)\) does not exist there. On the interval \((-\infty , 0)\), \(f\) has a critical number at \(x = \answer [given]{-1}\). On the interval \((0,\infty )\), \(f\) has a critical number at \(x = \answer [given]{\frac {1}{\sqrt {2}}}\). Mark the critical numbers \(x=-1\) and \(x=\frac {1}{\sqrt {2}}\) on your plot.

Does the function \(f\) have a critical number at \(x=-\frac {1}{\sqrt {2}}\)? Explain.

On the interval \((-\infty , -1)\), the derivative \(f'(x)=e^x(x+1)\) and has the same sign as the factor \((x+1)\), which is negative.

Therefore, \(f\) is decreasing on \((-\infty , -1)\).

On the interval \((-1, 0)\), the derivative \(f'(x)\) has the same sign as the factor \((x+1)\), which is positive.

Therefore, \(f\) is increasing on \((-1, 0)\).

On the interval \(\left (0, \frac {1}{\sqrt {2}}\right )\), the derivative \(f'(x)\) does not change the sign.

Since \(f'\left (\frac {1}{4}\right )= \frac {1}{16}-\frac {1}{2}<0\), it follows that \(f\) is decreasing on \(\left (0, \frac {1}{\sqrt {2}}\right )\).

On the interval \(\left (\frac {1}{\sqrt {2}}, \infty \right )\), the derivative \(f'(x)\) does not change the sign. Since \(f'(1)= 4\left (1-\frac {1}{2}\right )>0\), \(f\) is increasing on \(\left (\frac {1}{\sqrt {2}}, \infty \right )\).

The second derivative of \(f\) on \((0, \infty )\) is

On \((-\infty , 0)\), \(f''\) has one zero, namely \(x = \answer [given]{-2}\). The sign of \(f''\) changes from negative to positive] at this point.

On \((0, \infty )\), \(f''\) has one zero, namely \(x = \answer [given]{\frac {1}{\sqrt {6}}}\). The sign of \(f''\) changes from negative to positive] through this point.

Since all of this behavior as described above occurs on the interval \([-5,2]\), we now have a complete sketch of \(y=f(x)\) on this interval, see the figure below.