In the last section of the course, I want to pull together all of the techniques I have shown in order to develop a holistic way of describing functions. The first holistic problem is drawing graphs of function. Curve sketching is attempting to draw functions (without computer assistance) based on the properties derived using algebra and calculus.
Before I get to the general process of curve sketching, I need to talk about interpretation of derivatives. I have shown in the previous lectures that the first derivative gives information about increase and decrease. If \(f^\prime(x)\) is positive, the function is increasing, and if \(f^\prime(x)\) is negative, the function is decreasing. If \(f^\prime(x) = 0\text{,}\) then there may be a maximum or minimum.
There is a similar interpretation for the second derivative. The first derivative is slope, so the second derivative is the rate of change of slope. That measures the concavity of the graph: whether it is curving upwards or downwards. These shapes are called concave up and concave down, respectively. If \(f^{\prime
\prime}(x)\) is positive, then the function is concave up (curving updwards) and if \(f^{\prime \prime}(x)\) is negative, the function is concave down (curving downards). Where \(f^{\prime \prime}(x) = 0\) there might be a change in concavity; these points are called inflection points.
When I sketch a graph, I will consider all of the following eight attributes of the graph.
Domain. I determine domain by looking for domain restrctions such as division by zero.
Range. Range is still difficult, even with calculus, However, once I know minima, maxima, increase and decrease, I can often figure out the range. Therefore, even though I’ve listed it here, range is often the very last step.
Continuity. For standard functions, I get continuity for free on the domain. For piecewise functions, I have to do some work to investigate continuity at the cross-over points.
Intercepts. I evaluate \(f(0)\) to calculate the \(y\)-intercept, provided that \(x=0\) is in the domain. For the \(x\)-intercepts, I try to solve \(f(x) = 0\text{.}\)
Symmetry. I look for odd (\(f(-x) = -f(x)\)) or even (\(f(-x) = f(x)\)) symmetry, as well as periodicity (\(f(x+p) = f(x)\)).
Limits and Asymptotes. I look at limits near undefined points. If these limits are infinite, there are vertical asymptotes. I also look at limits at \(\pm \infty\text{.}\) If these limits are finite, there are horizontal asymptotes.
First Derivative. I calculate the first derivative and solve \(f^\prime(x) =0\) to find the critical points. I classify these points to find the minima or maxima. I also get the intervals of increase or decrease (which often let me determine the range, as mentioned before).
Second Derivative. I calculate the second derivative and solve \(f^{\prime \prime}(x) = 0\) to find the inflection points. I also get the intervals of concavity from the sign of the second derivative.
Subsection11.1.2Examples
Example11.1.1.
Consider this function: \(f(x) = x \ln x\text{.}\) I’ll go through the eight properties for sketching the graph of the function.
The domain is \(x > 0\text{,}\) due to the logarithm.
The range is unclear from first impressions.
The function is continuous on its domain.
There is no symmetry.
There is no \(y\) intercept. The \(x\) intercept is \((1,0)\text{.}\)
The only boundary point of the domain is \(x=0\text{.}\) The limit as \(x=0\) is calcluate by the limit approacing the undefined point (on the positive side only).
Since this limit is no infinite, there are no vertical asymptotes. The limit as \(x \rightarrow \infty\) is \(\infty\text{,}\) so there are no horizontal asymptotes either.
I look at the first derivative: \(f^\prime(x) = \ln x +
1\text{,}\) so \(f^\prime(x) = 0\) implies \(x =
\frac{1}{e}\text{.}\) The point \(\left( \frac{1}{e},
\frac{-1}{e} \right)\) is a potential extrema. I’ll go throught the process of looking at the intervals.
The point \(\left( \frac{1}{e}, \frac{-1}{e}
\right)\) is a local minimum.
I look at the second derivative: \(f^{\prime
\prime}(x) = \frac{1}{x}\text{.}\) The second derivative is always positive, so the function is concave up on \((0,
\infty)\text{,}\) which is the entire domain.
At the end of the process, I can return to the question of range. There is a minimum at \(x = \frac{1}{3}\text{,}\) with \(y\) value \(\frac{-1}{e}\text{.}\) Since this is a minimum and the fucntion grows to infinity, the range must be \(\left[ \frac{-1}{e}, \infty \right)\text{.}\)Figure 11.1.2 shows the resulting graph.
Figure11.1.2.\(f(x) = x \ln x\)
Example11.1.3.
Consider this function: \(f(x) = xe^{-x^2}\text{.}\) I’ll go through the eight steps for sketching the graph.
The domain is all of \(\RR\) since neither polynomials or exponentials have domain restrictions.
The range is uncertain at first glance.
The function is continuous.
\(f(0) = 0\) and that is the only intercept.
The function is odd, since \(f(-x) = -xe^{-(-x)^2} =
-xe^{-x^2} = - (x)\text{.}\)
There are no undefined points, so there are no vertical asymptotes. The limits as \(x \rightarrow \pm
\infty\) are both \(0\text{,}\) so there are horizontal asymptotes \(y=0\) in both the positive and negative directions.
The first derivative is \(f^\prime = e^{-x^2}
-2x^2e^{-x^2}\text{.}\) This vanishes when \(x= \pm
\frac{1}{\sqrt{2}}\text{.}\) I’lll look at the intervals for these critical point.
The point \(\left( \frac{1}{\sqrt{2}},
\frac{1}{\sqrt{2e}} \right)\) is a local maximum and the point \(\left( \frac{-1}{\sqrt{2}},
\frac{-1}{\sqrt{2e}} \right)\) is a local minimum. This calculation also gives the range, since these extrema are also global extrema. The range is \(\left[
-\frac{1}{\sqrt{2e}}, \frac{1}{\sqrt{2e}} \right]\text{.}\)
The second derivative is \(f^{\prime\prime} =
-6xe^{-x^2} + 4x^3e^{-x^2}\text{.}\) This vanishes when \(x
= \pm \sqrt{\frac{3}{2}}\text{.}\) which gives inflection points \(x=0\) and \(\left( \sqrt{\frac{3}{2}},
\sqrt{\frac{3}{2e^3}} \right)\) and \(\left(
-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2e^3}}
\right)\text{.}\) I’ll look at the interval to determine the concavity.
At the end of the process, we can reconsider range. There is a maximum at \(x = \frac{1}{\sqrt{2}}\text{,}\) with \(y\) value \(\frac{e^{-\frac{1}{2}}}{\sqrt{2}}\text{.}\) The function grows to this value and then decays to zero. The behaviour for the minimum is the same below the axis, due to the odd symmetry. Therefore, the range must be \(\left(
-\frac{e^{-\frac{1}{2}}}{\sqrt{2}},
\frac{e^{-\frac{1}{2}}}{\sqrt{2}} \right)\text{.}\)Figure 11.1.4 shows the resulting graph.
Figure11.1.4.\(f(x) = xe^{-x^2}\)
Example11.1.5.
Consider the quartic polynomial \(f(x) =
x^4+8x^3-270x^2+10\text{.}\) I’ll go through the standard eight steps for sketching the graph.
The domain is all of \(\RR\text{.}\)
The range is bounded below, since is it a quartic with a positive leading coefficient. However, that minimum is not obvious at first glance.
The function is continuous.
\((0,10)\) is the \(y\)-intercept. \(f\) has four roots, but calculating them is difficult. A computer approximation shows roots near \(x\) values of \(0.2\text{,}\)\(-0.2\text{,}\)\(-21\) and \(13\text{.}\)
There are no undefined points, so there are no vertical asymptotes. The limit at \(\pm \infty\) is \(\infty\text{,}\) so there are no horizontal asymptotes.
The first derivative is \(f^\prime (x) = 4x^3 + 24x^2 -
540x\text{,}\) which has roots of \(0\text{,}\)\(9\) and \(-15\text{.}\) The critical points are \((0, 1)\text{,}\)\((9, -9476)\) and \((-15, -37124)\text{.}\) I’ll look at the intervals to classify the critical points.
\((0,1)\) is the only maximum and the points \((-15,
-37124)\) and \((9, -9476)\) are mimima.
The second derivative is \(f^{\prime \prime}(x) = 12x^2
+ 48 x - 540\text{,}\) which has roots at \(x\) values \(5\) and \(-9\text{.}\) That gives possible inflection points at \((5, -5124)\) and \((-9, -21140)\text{.}\) I’ll look at the intervals to determine the concavity.
I said the range was boudned below. Now that I know the \(x\) coordinates of the two minima, I just need to which one has the least value. I asked a computer for these calculations. When \(x = -15\text{,}\) the function value is \(-37115\text{.}\) When \(x = 9\text{,}\) the function value is \(-9467\text{.}\) The first minimum is lower, so the range is \((-37114, \infty)\text{.}\)Figure 11.1.6 shows the resulting graph.
