Much can be done to sketch the approximate graph of a function without calculus, in fact I strongly encourage you to rely mostly on your pre-calculus skills to sketch graphs.
But by adding the derivative and second derivative to our toolbox, we can determine the exact locations of maxima and minima (collectively "extrema") of a function, and more. First a little terminology:
A global maximum or minimum might sometimes be called "local." Think of it like this Venn diagram. Global maxima and minima are a subset of local maxima and minima. On a given interval of the domain of a function, there can be only one global (or absolute) max. and min. — if either exists.
Remember that the derivative, f'(x), of the function f(x) gives the slope of f(x) at any differentiable point in its domain.
In the middle case, where the slope is zero, the function graph is neither rising nor falling, but is (usually) a relative maximum or minimum value. Fermat's theorem sums this up:
Fermat's theorem says that if a function has a local maximum or minimum (which could be global), then the derivative at that point is zero.
Referring to the definition of the local maximum above, we see that if a maximum value lies at x = c, then f(x) must be larger than some other value of the function f(x + h), where h can be positive or negative. So:
We can divide both sides of this inequality by h to make this look like a derivative, then take the limit as h → 0^{+} from the right:
Now we have assumed that f'(c) exists, so the limit from the right must equal the limit in general:
Plugging f'(c) into the last inequality, we get:
Now we could also do the same proof using the limit from the left, which would lead us to the inequality f'(c) ≥ 0. The only way that both inequalities can be true is for f'(c) to be equal to zero, so we have proved Fermat's theorem.
Go back and look at the Fermat's theorem box above and make sure to understand that the existence of a zero in the derivative function does not guarantee the existence of a relative or global max. or min. there. The graph of y = x^{3}, for example, has a horizontal tangent at x = 0, but it has no maxima or minima at all.
Here's a summary of the relationships of the first derivative of a function to maxima and minima (either local or global) in its graph:
It's very important when working these problems, not to forget your pre-calculus skills, the algebra skills you learned prior to calculus. Calculus won't replace the skills you already have, it will just enhance what you can learn about the graph of a function.
Don't forget your pre-calculus curve-sketching skills. Most of curve sketching does not require (and doesn't benefit from) calculus.
We begin by noting that this function can be factored by grouping:
The leading term is negative, so we know this function grows without bound on the left as x→ -∞, and tends to -∞ as x→ ∞ (i.e. goes up on the left & down on the right). We also know, from our familiarity with the sigmoidal (sideways-S) shape of the graphs of cubic functions, that the left-most extremum (which lies between x= -2 and x= 2, should be the local minimum and the right-most (between x=2 and x=4) the maximum. Here's a schematic picture:
Now to find the exact positions of the min. and max. using the derivative:
Here is the calculated graph of f(x). Hopefully you can see that by augmenting your pre-calculus curve sketching skills with calculus, you can learn a little more about the graph of a function.
But at the same time, don't forget that you could already say a lot about this graph with just those pre-calculus skills.
In the next example, we'll use that to find the solution to an optimization problem that, until now, you could probably only have solved by estimation with a computer or calculator.
Why do I need to learn to sketch curves if I can always just graph them on a computer?
There is no guarantee that the person who programs a calculator or computer (e.g. enters the function or data) does so without error. My hope for you is that you can be the person in the room who says, "wait a minute ... that can't be right ... " People make mistakes (and always will), so don't trust blindly in what your computer or calculator says. Be in a position to check it out, maybe not in great detail, but at least qualitatively, by applying your most excellent curve sketching skills.
Here is a schematic of the container. The top and bottom are circles of area A = πr^{2} and the side piece is a rectangle with area equal to the height (h) of the cylinder times its circumference, 2πr.
The surface area gives us the amount of material needed to build the cylinder. We'll use cm^{2} as the unit of area noting that 1 liter is 1000 cm^{3}.
The surface area is a function of two variables, r and h, but we can use the known volume to substitute for h as a function of r:
Now we have a single-variable equation for the surface area of the can:
Now we assume, because of the way we constructed this formula, that it has a maximum value. Now is the time to take out your calculator to confirm that it does. Take the derivative of A(r) with respect to r and set it equal to zero:
Solve for r:
The figure below shows a comparison between a polynomial function and its derivative. I've chosen the 6^{th} degree polynomial function f(x) = x^{6} - 3x^{5} + 6x^{3} - 3x^{2} - 3x + 2 for this comparison because it contains a variety of features, including two minima (one a global minimum), a local maximum and an inflection point.
Inflection point — a point in a graph at which the curvature changes between concave-upward and concave-downward.
Remember that the derivative gives the slope of a function at any point in its domain. Therefore when the slope of a function is negative, like the region between -∞ and -1 for f(x), the graph of the derivative should lie below y = 0, which it does (red curve). Follow this trend across the graph, through all of its regions of negative and positive slope. Wherever the slope of f(x) is increasing the graph of the derivative is positive; likewise for decreasing regions.
Next, sometimes the derivative graph passes through zero. That means f(x) has zero slope at those points. Zero is the slope of a horizontal tangent, which occurs at extrema or critical points. We always have to be careful of exceptions, however. In fact, this function has a zero in its derivative at x = 1 with no corresponding maximum or minimum there.
We can use the first derivative to determine whether a particular extreme value is a maximum or a minimum:
On both sides of x = 1 the slope of f(x) is negative except at x = 1, where it is zero. That means it can't be a maximum or a minimum (a case where the reverse of Fermat's theorem wouldn't work). The slope of the function must change sign on either side of a maximum or a minimum.
Now compare our function with its second derivative. The second derivative describes the change in the slope of a function - whether the slope of the function is increasing or decreasing. The second derivative couldn't be easier: It's just the derivative of the derivative:
The second derivative of a function is the derivative of its derivative.
There is a bit of extra notation there ( ↑ ). It's easiest, and very common, to use the compact notation f''(x) for the second derivative. The third notation is a little illogical (we're not taking a derivative with respect to
"x^{2}", but it's very commonly used, so you should get used to it. The third is due to Euler; it's compact and descriptive, but only used sometimes.
The second derivative is a good indicator of the curvature of a function.
In this comparison, notice that every zero (f''(x) = 0) in the graph of the second derivative matches an inflection point in the graph of f(x).
The zeros in f'(x) all align with inflection points. In particular, the zero at x = 1 confirms the inflection point we already knew.
It is possible, however, that a zero in the 2^{nd} derivative of a function does not indicate an inflection point. For example, the function g(x) = x^{4} has a zero at x = 0 in its second derivative, g''(x) = 12x^{2}, but that point is actually the global minimum of that function, with concave-up curvature on either side. We must be cautious when basing conclusions about inflection points on the second derivative.
Finally, the second derivative can supply more information about whether an extremum (singular of extrema) is a relative maximum or minimum:
For an extremum at a point c in the domain of f(x),
This is often referred to as the second derivative test.
Now we have another tool in our toolbox for understanding the shape of a graph: Perform the usual algebraic analysis, then use the first and second derivatives to find extrema and inflection points.
Some of the implications of the first and second derivatives on curve shape are summarized in the table below.
First, we note the the numerator is quadratic, with a negative discriminant, therefore the function has no real roots and its graph doesn't cross the x-axis:
The degree of the numerator (2) exceeds the degree of the denominator (1) by one, so the function has a slant asymptote. We find its equation by polynomial division, then note that the denominator has a zero at x = 2, therefore the function has a vertical asymptote there.
Now we find the extrema by taking the first derivative of f(x), then setting it equal to zero and solving for x.
Two zeros result, one at x = 0 and the other at x = 4. Now we take the derivative of the derivative, f''(x), to look for inflection points (f''(x) = 0), and to test whether our extrema are minima or maxima. This time I won't take up the space to calculate the second derivative of f(x), I'll just give the answer. You, however, should confirm that the answer is correct!
Looking back at our table of reminders, we see that if the second derivative is negative at an extreme value of the function domain, then there's a relative maximum there, and the opposite is true for a minimum. Notice that both the value of f'(x) and f''(x) are undefined at x = 2, as we would expect if there is a vertical asymptote at x = 2.
Here's the graph of our function. It has a slant asymptote (y = x) and a vertical asymptote at x = 2. The function is discontinuous and therefore not differentiable at x = 2. It has no holes. It has a relative maximum at x = 0 and a relative minimum at x = 4.
The calculus we did is completely consistent with what we discovered about this function using only our algebra toolkit, and it extends that technique by allowing us to find the exact location of the relative maximum and minimum.
The numerator is factorable, with real roots at x = 3 and x = -2, so we know that the graph crosses the x-axis at those (and only those) points. We can also easily see that the denominator is zero for x = 1, so there's a vertical asymptote at x = 1. There are no holes in the graph because there are no binomials (a + b) common to both the numerator and denominator.
It's also easy at this point to notice that the function crosses the y-axis at f(0) = 6:
The y-intercept is often very easy to calculate, so don't ignore it. Now we note the the degree of the denominator is one less than that of the numerator, so we expect a slanted asymptote, and we find its equation by polynomial division:
Now to find any maxima or minima, if they exist, we find the first derivative, f'(x), and set it equal to zero:
Setting the derivative equal to zero (below) , and solving by completing the square shows us that there are no real solutions, thus there are no places on the curve with a horizontal tangent slope — no maxima or minima at all:
Here is the graph of f(x).
This example is another good reminder not to forget about those precalculus skills. We found the y-intercept, the equation of the slant asymptote and the roots using algebra, and the facts they revealed basically established the shape of the graph. It's also worth thinking about the relative simplicity of the function — it just cant have a tremendous number of "wiggles" because its highest power is 2.
We used the first derivative to establish that there were no maxima or minima, sealing the deal on the shape of the graph; without any turning points and the other constraints we've stacked up, we've got all of the important features needed to sketch the graph. Absence of maxima or minima allows us to conclude that this function does not cross the slant asymptote for x near zero. If it did, that would create a "bump".
First, try to factor such a function; it could make life easier. This function can't be factored by grouping, but the rational root theorem, with a guess of x = 1 gets us to a factorable quadratic:
So we know that the graph crosses the x-axis at x = 1, -2 and 3. That's useful information.
Now the first derivative of f(x) is f'(x) = 3x^{2} - 4x - 5. Setting this to zero and finding the roots by completing the square (don't forget that!) gives two real roots:
So there is one maximum and one minimum (think about it, there can't be two maxima or two minima without the opposite intervening), but we don't know which is which. We can figure it out in two ways.
First, if the derivative is increasing on the left side of an extremum and decreasing on the right, the extremum is a maximum. The square root of 19 is about 4.5, so our extrema lie at about x = +2 and x = -2/3. Now f'(-1) is positive and f'(0) is negative, so the leftmost extremum is a maximum. Then f'(1) is negative and f'(3) is positive, so the other extremum is a minimum.
Second, we can apply the second derivative test. We have:
The two methods are in agreement.
← The function looks like this, and all of our discoveries, roots, y-intercept, and the relative maximum and relative minimum, are in the right place.
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Absolute value functions are tricky because they always have sharp points, like at the bottom of the V-shaped graph of y = |x|. These are called cusps.
A cusp is a point in a graph where two branches meet, but the tangents of each branch are not equal.
In order to analyze this graph, we'll have to take the derivative of an absolute-value function, and that's a little tricky. To do it, we need to define the absolute value of a number as the square root of the square of that
number. That makes sense: If we square a negative number, the result is positive, then we recover the number (without the negative sign) by taking the square root. The definition is in the box below.
According to the differentiation formula, f'(x) is:
Then if we set the derivative equal to zero, we get three solutions
Two of the solutions, x = ±1 are cusps. While the function is continuous at these points (you wouldn't have to lift your pencil to trace it), it is not differentiable.
We have to use our knowledge of functions in order to puzzle out the graph of f(x). If what's inside of the absolute value function has real zeros, those will be cusps, sharp turnarounds in the graph. In the function f(x) = |1-x^{2}|, there are zeros when x = ±1. If we rewrite the inside part as -x^{2} + 1, we see that it's just the parabola y = x^{2} flipped upside down and translated upward by one unit. So we'd expect the graph on the right above. The function y = 1-x^{2} is traced out, too, so you can make the
comparison. The other zero of the derivative, x = 0 is an extremum. We can use the second derivative to determine whether it's a minimum or maximum, though the absolute value signs force it to be a relative maximum anyway.
The second derivative of f(x) is a bit long, and you can work it out for yourself. It doesn't need to be simplified for you to just plug in x = 0 and determine that f''(0) = -2. That's less than zero, so x = 0 is a relative maximum.
We know the general shape of a root function, but recall that we can find odd roots of negative numbers, so this function is somewhat S-shaped.
Take the first derivative and notice that it is greater than zero for all x, and x can't be equal to zero. So this function is increasing throughout its domain, which includes zero, but its slope is undefined at zero.
The second derivative shows us that the function is concave-upward for x < 0 and concave downward for x > 0. The second derivative, too, is undefined at x = 0.
Here is the graph of the function →
It is continuous over its domain, (-∞, ∞), but non-differentiable at x = 0.
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