Curve Sketching

The process of curve sketching is to illustrate the graph of some function ${f(x)}$ by using ${f'(x)}$ and ${f''(x),}$ whether ${x}$ is positive or negative. To curve sketch, we state some key propositions up front:

Lemma. Let ${f}$ be a differentiable function. Then the following propositions are true:

1. If ${f' > 0,}$ then ${f}$ is increasing.
2. If ${f' < 0,}$ then ${f}$ is decreasing.
3. If ${f'' > 0,}$ then ${f'}$ is increasing.
4. If ${f'' < 0,}$ then ${f'}$ is decreasing.

Put differently, if a function ${f}$ is increasing, then its derivative's tangent lines are positive (i.e., becoming less negative). If a function ${f}$ is decreasing, then its derivative's tangent lines are negative (becoming less positive). We can visualize the principles above with a rough sketch:

Examining the sketches above, we see that the tangent lines on ${f}$ tend negative, corresponding to the graph of ${f'.}$ And with ${f',}$ the tangent lines tend positive, corresponding to the graph of ${f''.}$ With the graph of ${f'',}$ we see that the tangent lines tend negative, so a graph of ${f''',}$ the third derivative, would show a graph with a negative slope.

In the case where ${f'' > 0,}$ or when ${f'}$ is increasing, we say that ${f}$ is concave up. And when ${f'' < 0,}$ or when ${f'}$ is decreasing, we say that ${f}$ is concave down. In the sketches above, ${f}$ is concave down. Thus, we have the following propositions:

lemma. Let ${f}$ be a differentiable function. Then the following propositions are true:

1. If ${f'' < 0,}$ then ${f}$ is concave up.
2. If ${f'' < 0,}$ then ${f}$ is concave down.

We will denote these two propositions with the following notation:

notation. Given a function ${f,}$ if ${f^{\uparrow}}$ then ${f}$ is

concave up. If ${f^{\downarrow},}$ then ${f}$ is concave down. If ${f^{\leftrightarrow},}$ then ${f}$ is neither concave up nor concave down.

Let's apply these principles by considering the function ${f(x) = 3x - x^3.}$ Differentiating this function, we have:

Factoring the derivative:

We can draw several inferences from examining the derivative's equation. If ${x = 0,}$ then ${f'(x) = 3.}$ This is a positive number. More generally, the factor ${(1-x)(1+x)}$ is positive iff both ${(1-x)}$ and ${(1+x)}$ are greater than ${0.}$ Based on this fact, we can express this as an interval:

Hence, when ${-1 < x < 1,}$ then ${f'(x)}$ is positive. From the propositions we saw above, we infer that:

These aren't the only inferences we can draw. Looking back at the term ${(1-x)(1+x),}$ we see that it's negative if either of the terms are negative. When is ${(1-x)}$ negative? Again we solve for ${x}$ as an inequality:

And ${(1+x) < 0}$ when:

Thus, when ${(x > 1) \lor (x < -1),}$ ${f'}$ is negative. Finally, we can determine when ${f'}$ is zero:

Hence, when ${(x = 1) \lor (x = -1),}$ then ${f'}$ is zero. We've gathered quite a few inferences just from the derivative:

1. ${-1 < x < 1 \nc f' > 0 \nc}$ ${f}$ is increasing.
2. ${x > 1 \nc f' < 0 \nc}$ ${f}$ is decreasing.
3. ${x < -1 \nc f' < 0 \nc}$ ${f}$ is decreasing.
4. ${x = 1 \nc f' = 0 \nc}$ ${f}$ is constant.
5. ${x = -1 \nc f' = 0 \nc}$ ${f}$ is constant.

Based on this information, we can sketch the graph as follows:

Notice the yellow dots above. These points are called turning points or inflection points. They're simply points where the derivative (i.e., the slope) changes its sign. Formally:

definition. If ${f'(x_0) = 0,}$ then ${x_0}$ is a critical point, and the number ${y = f(x_0)}$ is called a critical value.

From the definition above, we see that the critical points are ${(-1, 0)}$ and ${(1, 0).}$ If we substitue ${x = -1}$ and ${x = 1}$ to the original function, ${f(x) = 3x - x^3,}$ we get:

Accordingly, we have the critical points:

Plotting these points on a graph:

Now, from our earlier sketch:

we know that at the point ${x = -1,}$ we hit ${f' = 0.}$ Before that point (x < -1,) ${f}$ is decreasing, and after that point ${(x > -1,)}$ ${f}$ is increasing. We also know that at the point ${x = 1,}$ we hit ${f' = 0,}$ but before that point (${x < 1}$) ${f}$ is increasing, and after that point, ${(x > 1),}$ ${f}$ is decreasing. Thus, we know that the graph along these points looks like:

We now continue filling in the gaps. First, it's always a good idea to step away from calculus for a moment and revisit the original function, applying techniques we learned from basic algebra and precalculus. The original function is ${f(x) = 3x - x^3.}$ This tells us that the function's graph has the property of crossing the point ${(0,0).}$ We confirm this by substituting ${x = 0:}$

So, we place that point as well:

Based on these facts alone, we have a wide range of possibilities for what this function might look like:

The goal now is to just keep drawing inferences to narrow down the possibilities. Looking at the original function, ${f(x) = 3x - x^3,}$ the function's equation consists of odd powers, so we know that this ${f}$ is an odd function. And since ${f}$ is an odd function, we know that the graph of ${f}$ is symmetric. Everything that's done on the right is done on the left.

Knowing this, the next step is to consider ${f}$'s extremeties.¹ By this we mean answering the following questions:

1. What happens when ${x \to \infty^{+}?}$
2. What happens when ${x \to \infty^{-}?}$

Let's consider the first question, when ${x}$ goes to positive infinity. When ${x}$ is a massive term, then the ${3x}$ term in ${3x - x^3}$ is negligible. This implies: When ${x}$ goes to positive infinity, then ${f(x) = 3x - x^3}$ behaves like ${f(x) = -x^3,}$ which goes to negative infinity. Symbolically:

Next, when ${x}$ goes to negative infinity, then the ${x^3}$ term becomes negligble, and ${f(x) = 3x - x^3}$ behaves like ${f(x) = 3x,}$ which tends towards positive infinity.

This tells us that with ${f(x) = 3x - x^3,}$ the part of the graph before the first critical point is pointing up, and the part of the graph after the second critical point is pointing down:

At this point, our sketch is getting pretty accurate. All that's left to do is to decorate. In other words, add more detail. We can do so by considering the second derivative:

Examining the expression ${-6x,}$ we know that:

This tells us that:

This tells us that where ${x < 0}$, ${f}$ is concave down, and where ${f > 0,}$ ${f}$ is concave up:

That's about as complete of a sketch as we can get. Here's the actual graph of ${f(x) = 3x - x^3:}$

That was a pretty accurate sketch. Importantly, the point ${(0,0)}$ is where the graph changes from concave up to concave down. We know this because ${f''(0) = 0:}$

Because of this special property, this point is called the inflection point. Let's consider another example. Let's say we want to sketch the graph of:

The derivative of ${f:}$

Looking at this derivative, we know that ${f'(x)}$ can never be ${0.}$ In other words, ${f'(x) \neq 0.}$ This in turn means that there are only two cases for ${f':}$

1. ${f' > 0}$ or
2. ${f' < 0.}$

This tells us that there are no critical points. Why? Because of the definition of a critical point: A critical point ${x_0}$ exists if, and only if, ${f'(x_0) = 0.}$ And since ${f'(x)}$ can never be ${0,}$ there are no points such that ${f'(x_0) = 0.}$

Whenever we encounter a situation like this — where we still don't have enough information after applying calculus — we want to go back to the basics; recall basic algebra and precalculus. Looking at the function:

We see that where ${x = -2,}$ ${f}$ is undefined:

From this deduction, we have the following:

By plotting ${x = -2,}$ we effectively impose a border. The function ${f}$ can never touch that line. The next step, is to answer the following:

What happens when ${f}$ gets very close to ${x = -2?}$

To answer this question, we keep in mind that there are two possibilities for ${f}$ approaching ${x = -2:}$

1. ${f}$ approaches ${x = -2}$ from the right, and
2. ${f}$ approaches ${x = -2}$ from the left.

When ${f}$ approaches ${x = -2}$ from the right, we write:

And when ${f}$ approaches ${x = -2}$ from the left, we write:

Examining the function, we have the following inferences. When ${f}$ approaches ${x = -2}$ from the right, we get:

When we write ${(-2)^{+},}$ we mean that it's not quite ${-2.}$ Instead, it's just a little more than ${-2,}$ say ${-1.99999.}$ Similarly, when we write ${(0)^+,}$ we mean that it's just tiny bit more than ${0.}$ We can think of it this way:

In other words, it's some number just a little more than ${-1,}$ divided by some number that's just a little more than ${0.}$ This means we have a negative number divided by a very, very small positive number. This yields negative infinity:

Thus, when ${f}$ approaches ${x = -2}$ from the right, it goes towards negative infinity. Next, when ${f}$ approaches ${x = -2}$ from the left, we have:

which we can think of as:

Here, we have a negative number divided by a very, very small negative number. This yields positive infinity:

All together, our analysis yielded several conclusions. When ${f}$ approaches ${x = -2}$ from the right, it tends towards negative infinity. And when ${f}$ approaches ${x = -2}$ from the left, it tends towards positive infinity. Symbolically:

1. ${x \to (-2)^{+} \nc f(x) \to \infty^{-}}$
2. ${x \to (-2)^{-} \nc f(x) \to \infty^{+}}$

Because we have these inferences, we can next consider ${f}$'s extremities

1. What happens when ${x}$ is very large? (i.e., ${x \to \infty^{+}}$)
2. What happens when ${x}$ is very small? (i.e., ${x \to \infty^{-}}$)

For the first question, we evaluate the limit:

This limit is a little difficult to interpret as is, so we rewrite it by dividing the numerator and the denominator by ${x:}$

Looking at it this way, we can see that as ${x \to \infty^{+},}$ ${\dfrac{1}{x} \to 0}$ and ${\dfrac{2}{x} \to 0.}$ Thus:

Just looking at the rewritten expression, the analysis doesn't change when ${x}$ tends to negative infinity. We still tend to ${1:}$

Thus, we have the following inferences:

1. ${x \to \infty^{+} \nc f(x) \to 1.}$
2. ${x \to \infty^{-} \nc f(x) \to 1.}$

We can abstract both inferences above as:

So, putting all of our inferences together, we have:

1. Proposition 1: ${x \to (-2)^{+} \nc f(x) \to \infty^{-}}$
2. Proposition 2: ${x \to (-2)^{-} \nc f(x) \to \infty^{+}}$
3. Proposition 3: ${f(\infty^{\pm}) = 1}$

From these propositions, we know that there are two special asymptotes for the graph of ${f,}$ namely: ${x = -2}$ and ${y = 1.}$ On a graph:

Next, from proposition 1, we know that as we approach ${-2}$ from the right, we plunge towards negative infinity:

And from proposition ${2,}$ we know that as we approach ${-2}$ from the left, we fly towards positive infinity:

Finally, from proposition ${3,}$ we know that when for very small ${x,}$ we're tending towards ${y = 1,}$ and for very large ${x,}$ we're also tending towards ${y = 1:}$

Now we have to consider the missing portions above. The key question is, does this graph ever dip into ${y = 1}$ and come back up? The answer is

no. Why? Because there are no critical points. If ${f}$ did dip below and come back up, we would have a critical point. But we know for a fact that there are not. Since ${f' \neq 0,}$ we know that we can never have a horizontal tangent line, which means that the the graph of ${f}$ can never dip down and come back up. As such, the graph of ${f}$ should look like:

At this point, we're essentially done, but it's worth doing a few double-checks. To do so, we consider the derivatives. First, we rewrite the function's expression:

The first derivative is now easier to compute:

Examining the first derivative, we easily see that ${f}$ is just a hyperbola, so we're correct in sketching a graph consisting of two arms. The first derivative also tells us that:

${f}$ is increasing on the intervals: ${\infty^- < x < -2,}$ and ${-2 < x < \infty^+}$

It's importat that we specify the interval, because it is not true ${f}$ is increasing. It's increasing only at a specific point — where ${x \neq -2.}$ Next, we consider the second derivative:

Looking at the second derivative, we can draw the following inferences:

1. ${-2 < x < +\infty \iff f''(x) < 0}$ (concave down)
2. ${-\infty < x < -2 \iff f''(x) > 0}$ (concave up)

The inferences above tell us that there's no "wiggle" in ${f}$'s graph. In other words, it rules out the possibility of one of the arms doing something like this:

And with that, we're done. Here's the actual graph of ${f:}$

Having gone through the examples, here's a general strategy for sketching:

• For basic sketching:
  1. Plot any discontinuities (especially infinite discontinuities).
  2. Plot the endpoints or where ${x \to \pm \infty.}$
  3. Plot any immediate points.
• For first-derivative-informed sketching:
  1. Solve for ${x,}$ where ${f'(x) = 0.}$
  2. Plot the resulting critical points and values, if any.
  3. Determine whether ${f' < 0}$ or ${f' > 0}$ on each interval between the critical points or discontinuities.
• Second-derivative-informed sketching:
  1. Determine whether ${f'' > 0}$ or ${f'' < 0}$ (concave up or concave down).
  2. Solve for ${x,}$ where ${f''(x) = 0.}$
  3. Plot the resulting inflection points, if any.