Day 21, Mat128: Calculus A

• Any symmetry? Any periodicity? Do we recognize the class of this function, and do we already have a good idea of what it's going to look like? These may allow us to take some shortcuts.

• We move to the domain. Then we look for discontinuities -- where the function misbehaves.
  • Vertical asymptotes
  • Jump discontinuities
  • Removable discontinuities don't really affect the shape of the curve since they are holes. (These are like the indeterminate limit treated in the Preview of section 2.8 for L'Hôpital's rule)

• Intercepts of the x- and y-axes (if any):
  • \(y\)-intercept is where \(x=0\) so that \(y=f(0).\) There can be at most one \(y\)-intercept, and it is usually easy to determine. This can help with determining what the plot window should be, at least the \(y\)-values.
  • \(x\)-intercepts are where \(y=0\) so it involves solving the equation \(0=f(x).\) This may or may not be easy to determine, and will help determine what portion of the plot window we want to show -- along with zeros of the first and second derivatives.
• End behavior as \(\vert x\vert\to\infty.\)
  • If it has horizontal asymptotes on either end, its growth or decline levels off.
  • If it has no horizontal asymptotes, growth or decline continues indefinitely or it oscillates.

The first derivative tells where the function is increasing and where it is decreasing. The only places where the the function can switch from increasing to decreasing, or vice versa, is either about discontinuities or about first order critical numbers. In any case, it is useful to know

• Where \(f'(x)=0,\) points where the graph has horizontal tangents. These points are candidates for local maximum points or local minimum points.
• Where \(f'(x)\) is not defined or where either there is a sharp corner or the original function has a discontinuity. The former case (corners) are candidates for local maximum points or local minimum points.
• We examine the sign of the first derivative between these points. Where positive the curve is increasing; where negative, decreasing.

The second derivative tells about the concavity of the function. The only places where the function can switch concavity is about either discontinuities or about second order critical numbers. The second order critical numbers are candidates for inflection points.

In summary, most of the shape of a graph comes from the following information.

• Where \(f(x)=0\) and where \(f(x)\) is undefined, and the sign between those points.
• Where \(f'(x)=0\) and where \(f'(x)\) is undefined, and the sign between those points.
• Where \(f''(x)=0\) and where \(f''(x)\) is undefined, and the sign between those points.

So we'll frequently construct a table with that information. It gives a cartoon picture of a function. We also include the asymptotes, vertical and horizontal, and any other special points.

1. Consider the function \(f(x)=\frac{4}{x^2+2x-8}.\)

   • Function information:
     • Its \(y\)-intercept is \(y=f(0)=\frac{4}{-8}=-0.5.\)
     • Its \(x\)-intercepts are solutions to \(0=f(x)\rightarrow 0=4.\) There are no solutions to this so it never crosses the \(x\)-axis.
     • It has discontinuities at zero divides: \(0=x^2+2x-8=(x+4)(x-2)\rightarrow x=-4,x=2.\) These are vertical asymptotes.
     • As \(x\to\pm\infty\) \(f(x)\approx \frac{4}{x^2}\to 0.\) On either end it levels off to \(0\).
     • We will see that this function is, indeed, symmetric -- but about a line other than $x=0$, which makes it a little harder to see at first. But it should be clear that this is so, because the numerator is a constant, and the denominator is a quadratic, whose graph is symmetric (a parabola).
   • First derivative information: ("low d'high minus high d'low, over the denominator square they go!")

     \(f'(x)=\frac{0(x^2+2x-8)-(4)(2x+2)}{(x^2+2x-8)^2}=\frac{-8(x+1)}{(x^2+2x-8)^2}.\)

     This is zero for \(x=-1\) and it is not defined at the two vertical asymptotes.

     The sign changes only at $x=-1$: it is positive to the left of that, and negative to the right.

   • Second derivative information: This starts to be a little messy (so let technology do it), to get
     \(f''(x)=\frac{18(x^2+2x+4)}{(x^2+2x-8)^3}.\)

     The quadratic in the numerator has no real roots, so the second derivative is never zero.

     The denominator is zero at the vertical asymptote points, and changes sign as it passes them. It is positive to the left, negative between, and positive to the right.

   • Now we'll create the table to incorporate all this information.

     All zeros of $f$, $f'$, and $f''$ should appear, as well as vertical asymptotes, etc.

     $x$
     $f(x)$
     $f'(x)$
     $f''(x)$

   • To plot this function we should include values of \(x=-4,-1,2,\) and the \(y\)-values should fairly tight around \(0.\) Let us make the viewing window \(-6≤x≤4, -2≤y≤3.\)
     (Usually vertical and horizontal asymptotes are denoted as dashed, but I'm not sure how to do that with this particular software!:)

2. Consider the function \(f(x)=x\ln{x}.\)

   • Function information:
     • There are no symmetries or periodicities to take advantage of. The domain is $x \gt 0$ (because of the log function).
     • Its \(y\)-intercept doesn't exist.
     • Its \(x\)-intercepts are solutions to \(0=f(x).\) It has one at $x=1$, when the log is 0.
     • It has no discontinuities on its domain. Both functions making up the product are smooth on the domain.
     • As noted above, using L'Hôpital's rule, we can conclude that $\lim_{x\to 0}f(x)=0$.
     • As \(x\to\pm\infty\) \(f(x)\to \infty\) (since both $x$ and $\ln(x)$ do, their product does as well -- this is not indeterminate).
   • First derivative information: \(f'(x)=1+\ln(x).\) This is zero for \(x=e^{-1}\) -- by the undoing property of $\ln(x)$ and $e^x$: \[ 1 + \ln(x) = 0 \iff \ln(x) = -1 \] Raise both sides to the power $e$, and we have \[ e^{\ln(x)} = e^{-1} \iff x = e^{-1}. \]

   • Second derivative information: This one isn't messy at all! \[ f''(x)=\frac{1}{x} \] which is never zero (and always positive) on the domain. This function is concave up everywhere.

   • Now we'll create the table to incorporate all this information.

     All zeros of $f$, $f'$, and $f''$ should appear, as well as vertical asymptotes, etc.

     $x$
     $f(x)$
     $f'(x)$
     $f''(x)$

   • To plot this function we should include values of \(x=0,1,\) and the \(y\)-values won't get any lower than \(-e^{-1}\) Let us make the viewing window \(-1, 2.5, -1, 1.5.\)