2.8: Curve Sketching

List the intervals on which the function shown is increasing or decreasing.

Answer:

[latex]f[/latex] is increasing on the intervals [0,1], [ 2,3] and [4,6].
[latex]f[/latex] is decreasing on [1,2] and [6,8].
On the interval [3,4] the function is not increasing or decreasing – it is constant.

The graph shows the height of a helicopter during a period of time. Sketch the graph of the upward velocity of the helicopter, [latex]\frac{dh}{dt}[/latex].

Answer:

Notice that the [latex]h(t)[/latex] has a local maximum when [latex]t = 2[/latex] and [latex]t = 5[/latex], and so [latex]v(2) = 0[/latex] and [latex]v(5) = 0[/latex]. Similarly, [latex]h(t)[/latex] has a local minimum when [latex]t = 3[/latex], so [latex]v(3) = 0[/latex].
When [latex]h[/latex] is increasing, [latex]v[/latex] is positive. When [latex]h[/latex] is decreasing, [latex]v[/latex] is negative.
Using this information, we can sketch a graph of [latex]v(t) = \frac{dh}{dt}[/latex].

The next theorem is almost the converse of the First Shape Theorem and explains the relationship between the values of the derivative and the graph of a function from a different perspective. It says that if we know something about the values of [latex]f'[/latex], then we can draw some conclusions about the shape of the graph of [latex]f[/latex].

Use information about the values of [latex]f'[/latex] to help graph [latex]f(x) = x^3 - 6x^2 + 9x + 1[/latex].
[latex]f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3)[/latex] so [latex]f'(x) = 0[/latex] only when [latex]x = 1[/latex] or [latex]x = 3[/latex]. [latex]f'[/latex] is a polynomial so it is always defined.

Answer:

The only critical numbers for [latex]f[/latex] are [latex]x = 1[/latex] and [latex]x = 3[/latex], and they divide the real number line into three intervals: [latex](-\infty, 1)[/latex], [latex](1,3)[/latex], and [latex](3, \infty)[/latex]. On each of these intervals, the function is either always increasing or always decreasing.
If [latex]x \lt 1[/latex], then [latex]f '(x) =[/latex] 3(negative number)(negative number) [latex]\gt 0[/latex] so [latex]f[/latex] is increasing.
If [latex]1 \lt x \lt 3[/latex], then [latex]f'(x) =[/latex] 3(positive number)(negative number) [latex]\lt 0[/latex] so f is decreasing.
If [latex]x \gt 3[/latex], then [latex]f'(x) =[/latex] 3(positive number)(positive number) [latex]\gt 0[/latex] so [latex]f[/latex] is increasing.
Even though we don’t know the value of [latex]f[/latex] anywhere yet, we do know a lot about the shape of the graph of [latex]f[/latex]: as we move from left to right along the [latex]x[/latex]-axis, the graph of [latex]f[/latex] increases until [latex]x = 1[/latex], then the graph decreases until [latex]x = 3[/latex], and then the graph increases again. The graph of [latex]f[/latex] makes “turns” when [latex]x = 1[/latex] and [latex]x = 3[/latex]; it has a local maximum at [latex]x = 1[/latex], and a local minimum at [latex]x = 3[/latex].

To plot the graph of [latex]f[/latex], we still need to evaluate [latex]f[/latex] at a few values of [latex]x[/latex], but only at a very few values. [latex]f(1) = 5[/latex], and (1,5) is a local maximum of [latex]f[/latex]. [latex]f(3) = 1[/latex], and (3,1) is a local minimum of [latex]f[/latex]. The resulting graph of [latex]f[/latex] is shown here.

Use information about the values of [latex]f''[/latex] to help determine the intervals on which the function [latex]f(x) = x^3 - 6x^2 + 9x + 1[/latex] is concave up and concave down.

Answer: For concavity, we need the second derivative: [latex]f'(x)=3x^2-12x+9[/latex], so [latex]f''(x)=6x-12[/latex].
To find possible inflection points, set the second derivative equal to zero. [latex]6x-12=0[/latex], so [latex]x = 2[/latex]. This divides the real number line into two intervals: [latex](-\infty,2)[/latex] and [latex](2,\infty)[/latex].
For [latex]x \lt 2[/latex], the second derivative is negative (for example, [latex]f''(0)=6(0)-12=-12[/latex]), so [latex]f[/latex] is concave down. For [latex]x \gt 2[/latex], the second derivative is positive, so [latex]f[/latex] is concave up.
We could have incorporated this concavity information when sketching the graph for the previous example, and indeed we can see the concavity reflected in the graph shown.

Use information about the values of [latex]f'[/latex] and [latex]f''[/latex] to help graph [latex]f(x)=x^{2/3}[/latex].
[latex]f'(x)=\frac{2}{3}x^{-1/3}[/latex]. This is undefined at [latex]x = 0[/latex].
[latex]f''(x)=-\frac{2}{9}x^{-4/3}[/latex]. This is also undefined at [latex]x = 0[/latex].
This creates two intervals: [latex]x \lt 0[/latex], and [latex]x \gt 0[/latex].

Answer:On the interval [latex]x \lt 0[/latex], we could test out a value like [latex]x = -1[/latex]: \[ f'(-1)=\frac{2}{3}(-1)^{-1/3}=-\frac{2}{3} \] and \[ f”(-1)=-\frac{2}{9}(-1)^{-4/3}=-\frac{2}{9}. \]
[latex]f'(x)[/latex] is negative and [latex]f''(x)[/latex] is negative, so we can conclude that the function is decreasing and concave down on this interval.
On the interval [latex]x \gt 0[/latex], we could test out a value like [latex]x = 1[/latex]: \[ f'(1)=\frac{2}{3}(1)^{-1/3}=\frac{2}{3} \] and \[ f”(1)=-\frac{2}{9}(1)^{-4/3}=-\frac{2}{9}. \]
[latex]f'(x)[/latex] is positive and [latex]f''(x)[/latex] is negative, so we can conclude that the function is increasing and concave down on this interval.
We can also calculate that [latex]f(0)=0[/latex], giving us a base point for the graph. Using this information, we can conclude the graph must look like this:

A company’s bank balance, [latex]B[/latex], in millions of dollars, [latex]t[/latex] weeks after releasing a new product is shown in the graph below. Sketch a graph of the marginal balance – the rate at which the bank balance was changing over time.

Answer:

Notice that since the tangent line will be horizontal at about [latex]t = 0.6[/latex] and [latex]t = 3.2[/latex], the derivative will be 0 at those points.
We can then identifying intervals on which the original function is increasing or decreasing. This will tell us when the derivative function is positive or negative.

Interval	[latex]B(t)[/latex]	[latex]B'(t)[/latex]
[latex]0 \lt t \lt 0.6[/latex]	Decreasing	Negative
[latex]0.6 \lt t \lt 3.2[/latex]	Increasing	Positive
[latex]t \gt 3.2[/latex]	Decreasing	Negative

There appear to be inflection points at about [latex]t =1.5[/latex] and [latex]t = 5.5[/latex]. At these points, the derivative will be changing from increasing to decreasing or vice versa, so the derivative will have a local max or min at those points.
Looking at the intervals of concavity:

Interval	[latex]B(t)[/latex]	[latex]B'(t)[/latex]
[latex]0 \lt t \lt 1.5[/latex]	Concave Up	Increasing
[latex]1.5 \lt t \lt 5.5[/latex]	Concave Down	Decreasing
[latex]t \gt 5.5[/latex]	Concave Up	Increasing

If we wanted a more accurate sketch of the derivative function, we could also estimate the derivative at a few points:

[latex]t[/latex]	[latex]B'(t)[/latex]
0	-10
1.5	3
6	-1

Now we can sketch the derivative. We know a few values for the derivative function, and on each interval we know the shape we need. We can use this to create a rough idea of what the graph should look like.

Smoothing this out gives us a good estimate for the graph of the derivative.