Curve Sketching

This placeholder page contains free typical curve sketching problems, each with a detailed solution, for you to practice. Curve sketching problems tie together many of the skills you've been developing, so are a favorite of instructors to put on exams since they test many things at once. They thus serve as excellent exam preparation, so practice here, where you can make and correct any small errors without penalty – and be ready for your exam!

Practicce Problem #1: 𝑓(𝑥) =𝑥3+1𝑥2

Consider the function 𝑓(𝑥) =𝑥3+1𝑥2.

(a)

Is 𝑓(𝑥) even, odd, or neither?

(b)

In which intervals is 𝑓(𝑥) concave up? concave down?

(c)

Sketch the curve 𝑦 =𝑓(𝑥).

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• Solution Summary
• Solution (a) Detail
• Solution (b) Detail
• Solution (c) Detail

• (a) Neither. (See detailed solution for explanation.)
• (b) Concave up for ( −∞,0) and (0,∞); never concave down.
• (c) See detailed solution.

Recall that for an even function, 𝑓( −𝑥) =𝑓(𝑥), while for an odd function 𝑓( −𝑥) = −𝑓(𝑥). We thus need to determine how 𝑓( −𝑥) compares to 𝑓(𝑥):

𝑓(−𝑥)=(−𝑥)3+1(−𝑥)2=−𝑥3+1𝑥2

If the function were even, 𝑓( −𝑥) would equal 𝑓(𝑥) =𝑥3+1𝑥2, which it does not.

If the function were odd, 𝑓( −𝑥) would equal −𝑓(𝑥) =−(𝑥3+1)𝑥2, which it does not.

Hence the function is neither even nor odd. ✓

To find where the function is concave up and concave down, we must investigate its second derivative, 𝑓″(𝑥):

𝑓′(𝑥)=1−2𝑥3𝑓″(𝑥)=0−(2)(−3)𝑥4=6𝑥4

Hence 𝑓″(𝑥) is positive for all 𝑥 (except at 𝑥 =0, where the function is undefined): the function is concave up for ( −∞,0) and (0,∞). It is never concave down. ✓

To sketch the curve, let's note first that the function is undefined at 𝑥 =0. Investigating what happens as we approach 𝑥 =0 from the left and from the right:
∙lim𝑥→0−[𝑥+1𝑥2]=0+lim𝑥→0−1𝑥2=∞
∙lim𝑥→0+[𝑥+1𝑥2]=0+lim𝑥→0+1𝑥2=∞
Hence we can sketch the first part of the curve as shown, remembering that we found in parts (b) and (c) that the curve is concave up everywhere (except at 𝑥 =0), and is increasing we approach the 𝑦-axis from the left, and decreasing as we leave the 𝑦-axis going to the right. Next, let's determine if there are any 𝑥-intercepts, which would occur when 𝑦 =𝑥3+1𝑥2 =0. The function is undefined for 𝑥 =0, so we only need to determine when (if) the numerator equals 0:

𝑥3+1=0𝑥3=−1𝑥=−1 Hence there is an 𝑥-intercept at 𝑥 = −1. We add this knowledge to our sketch as shown. Let's next consider what happens to 𝑓(𝑥) =𝑥 +1𝑥2 as 𝑥 becomes very large in the negative direction (𝑥 → −∞), and becomes very large in the positive direction (𝑥 →∞):
∙ As 𝑥 → −∞, 1𝑥2 →0, and so 𝑦 =𝑓(𝑥) becomes very close to 𝑦 =𝑥.
∙ As 𝑥 →∞, 1𝑥2 →0, and so again 𝑦 =𝑓(𝑥) becomes very close to 𝑦 =𝑥, shown as a dotted line in the figure below. Finally, we found in part (b) that 𝑓(𝑥) has a local minimum at 𝑥 =3√2: it is decreasing on the interval (0,3√2) and increasing on (3√2,∞) Putting the pieces together, we sketch this graph: For comparison, the actual graph is shown below; the second one is more zoomed-out.

Practice Problem #2: 𝑓(𝑥) =𝑥4/3 +4𝑥1/3

Consider the function 𝑓(𝑥) =𝑥4/3 +4𝑥1/3 on the interval −8 ≤𝑥 ≤8.

(a)

Find the coordinates of all points at which the tangent to the curve is a horizontal line.

(b)

Find the coordinates of all points at which the tangent to the curve is a vertical line.

(c)

Find the coordinates of the points at which the absolute maximum and absolute minimum occur.

(d)

For what values of 𝑥 is the function concave down?

(e)

Sketch the curve 𝑦 =𝑓(𝑥).

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• Solution Summary
• Solution (a) Detail
• Solution (b) Detail
• Solution (c) Detail
• Solution (d) Detail
• Solution (e) Detail

• (a) ( −1, −3)
• (b) (0,0)
• (c) Absolute minimum at ( −1, −3); absolute maximum at (8,24)
• (d) On the interval (0, 2).
• (e) See detailed solution.

To find the slope of the tangent line to the curve, we examine the function's derivative:

𝑓′(𝑥)=𝑑𝑑𝑥[𝑥43+4𝑥13]=43𝑥13+4(13)𝑥−23=43(𝑥+1)𝑥23

The tangent line is horizontal when the curve's slope is zero (𝑓′(𝑥) =0). We can see immediately that this occurs when 𝑥 = −1.

To fully specify the point where the curve has zero slope, we need to find the 𝑦-value when 𝑥 = −1: 𝑦=(−1)43+4(−1)13=1+4(−1)=−3 Hence the curve has a horizontal tangent at the point ( −1, −3). ✓

In part (a) we found that the function's derivative is 𝑓′(𝑥)=43(𝑥+1)𝑥23 The tangent line is vertical when the derivative does not exist (when it is infinite). We see immediately that this occurs when 𝑥 =0.

When 𝑥 =0, 𝑦 =𝑓(0) =0. So the curve has a vertical tangent line at the point (0,0). ✓

Recall from part (a) that the function's derivative is 𝑓′(𝑥)=43(𝑥+1)𝑥23 The critical points are thus 𝑐 = −1 and 𝑐 =0.

We thus need to examine the three regions (i) 𝑥 < −1; (ii) −1 <𝑥 <0; and (iii) 0 <𝑥.

∙ (i) For 𝑥 < −1, 𝑓′(𝑥) <0, since the numerator is negative while the denominator is positive.

∙ (ii) For −1 <𝑥 <0, 𝑓′(𝑥) >0, since the numerator is positive and the denominator is positive.

∙ (iii) For 𝑥 >0, 𝑓′(𝑥) >0, since again the numerator is positive and the denominator is positive. Hence there is a local minimum at 𝑥 = −1.

To find the absolute maximum and minimum, we must check the endpoints to see how they compare to the local minimum at 𝑥 = −1:

∙ 𝑓( −1) =( −1)43 +4( −1)13 =1 +4( −1) = −3
∙ 𝑓( −8) =( −8)43 +4( −8)13 =16 +4( −2) =8
∙ 𝑓(8) =843 +4(8)13 =16 +4(2) =24


Hence the absolute minimum is the point (-1 ,-3) and the absolute maximum is the point (8, 24). ✓

To determine concavity we examine the function's second derivative. We begin with our result from part (a) for the first derivative. Let's start with the equation we obtained right after taking the first derivative, before we did any simplifying, since taking the second derivative is then easier: 𝑓′(𝑥)=43𝑥13+43𝑥−23𝑓″(𝑥)=(43)(13)𝑥−23+(43)(−23)𝑥−53=49[𝑥−2𝑥53]

The numerator changes sign at 𝑥 =2, while the denominator changes sign at 𝑥 =0. We thus see that
∙ For 𝑥 <0, 𝑓″(𝑥) >0 since both the numerator and denominator are negative.
∙ For 0 <𝑥 <2, 𝑓″(𝑥) <0 since the numerator is negative while the denominator is positive.
∙ For 𝑥 >2, 𝑓″(𝑥) >0 since both the numerator and denominator are positive.
Hence the function is concave down on the interval (0, 2). ✓

Let's summarize what we've found in the earlier parts about the function on the interval [-8, 8]:
∙ The function's tangent is vertical at 𝑥 =0, so the graph itself is vertical there.
∙ The graph has an absolute minimum at (-1, -3).
∙ The graph has an absolute maximum at (8, 24).
∙ The graph is concave down on (0,2), and concave up on the other intervals.

We've captured these features in the figure shown. Filling in the rest of the graph can result in something like the figure below. A computer-generated graph of the function is shown below for comparison.

Practice Problem #3: 𝑦 =𝑥 +sin⁡𝑥

Consider the function defined by 𝑦 =𝑥 +sin⁡𝑥 for all 𝑥 such that −𝜋2 ≤𝑥 ≤3𝜋2.

(a)

Find the coordinates of all maximum and minimum points on the given interval. Justify your answers.

(b)

Find the coordinates of all points of inflection on the given interval. Justify your answers.

(c)

Sketch the graph of the function.

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• Solution Summary
• Solution (a) Detail
• Solution (b) Detail
• Solution (c) Detail

• (a) Maximum: (3𝜋2,3𝜋2−1); minimum (−𝜋2,−𝜋2−1)
• (b) (0,0) and (𝜋,𝜋)
• (c) See detailed solution.

To find where the maxima and minima lie, we must examine the derivative of the function:

𝑓′(𝑥)=𝑑𝑑𝑥[𝑥+sin⁡𝑥]=1+cos⁡𝑥 To find the critical points, we set 𝑓′(𝑐) =0:

𝑓′(𝑐)=01+cos⁡𝑐=0cos⁡𝑐=−1[Recall that −𝜋2≤𝑥≤3𝜋2]𝑐=𝜋

Our only critical point is at 𝑐 =𝜋. But notice that since the most negative cos⁡𝑥 can become is -1, 𝑓′(𝑥) =1 +cos⁡𝑥 is never negative. Hence 𝑥 =𝜋 is neither a maximum nor a minimum. (Thinking ahead to part (c), note however that 𝑓(𝑥) has a horizontal tangent at 𝑥 =𝜋.)

Let's consider the endpoints of the interval:
∙ 𝑓( −𝜋2) = −𝜋2 +sin⁡(−𝜋2) = −𝜋2 −1

∙ 𝑓(3𝜋2) =3𝜋2 +sin⁡(3𝜋2) =3𝜋2 −1


Hence the maximum point is (3𝜋2,3𝜋2−1), and the minimum point is (−𝜋2,−𝜋2−1) ✓

To find the point(s) of inflection, we must examine the second derivative. We begin with the first derivative that we found in part (a): 𝑓′(𝑥) =1 +cos⁡𝑥.

𝑓′(𝑥)=1+cos⁡𝑥𝑓″(𝑥)=−sin⁡𝑥 The points of inflection occur when 𝑓″(𝑥) =0:
−sin⁡𝑥=0

On our interval [−𝜋2,3𝜋2], −sin⁡𝑥 =0 at

𝑥=0,𝜋

We thus need to examine 𝑓″(𝑥) in the three regions: (i) −𝜋2 <𝑥 <0; (ii) 0 <𝑥 <𝜋; (iii) 𝜋 <𝑥 <3𝜋2:
∙ (i) For −𝜋2 <𝑥 <0, −sin⁡𝑥 >0.
∙ (ii) For 0 <𝑥 <𝜋, −sin⁡𝑥 <0.
∙ (iii) For 𝜋 <𝑥 <3𝜋2, −sin⁡𝑥 >0. Hence we see that both 𝑥 =0 and 𝑥 =𝜋 are points of inflection.

We need the 𝑦-value of those points in order to answer the question as asked:
∙ 𝑓(0) =0 +sin⁡0 =0
∙ 𝑓(𝜋) =𝜋 +sin⁡𝜋 =𝜋

The coordinates of the points of inflection are therefore (0, 0) and (𝜋,𝜋). ✓

Let's summarize what we know from the preceding parts of this question:
∙ The function has a minimum at (−𝜋2,−𝜋2−1).
∙ The function has maximum at (3𝜋2,3𝜋2−1).
∙ The function has a horizontal tangent at (𝜋,𝜋).
∙ The function is concave down on the interval (0,𝜋), and concave up on the other two intervals.

We've sketched these facts in this figure. Joining the pieces together leads to something like this figure. The figure below is a computer-generated graph of the function for comparison.

Practice Problem #4: 𝑓(𝑥) =cos2⁡𝑥 +2cos⁡𝑥

Consider the function 𝑓(𝑥) =cos2⁡𝑥 +2cos⁡𝑥 on the interval [0,2𝑝𝑖].

(a)

Find all values of 𝑥 in this at which 𝑓(𝑥) =0.

(b)

Find all values of 𝑥 in this period at which the function has a minimum. Justify your answer.

(c)

Over what intervals in this period is the curve concave up?

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• Solution Summary
• Solution (a) Detail
• Solution (b) Detail
• Solution (c) Detail

• (a) 𝜋2, 3𝜋2
• (b) 𝜋
• (c) (𝜋3,𝜋) and (𝜋,5𝜋3)

To determine when 𝑓(𝑥) =0, let's factor the expression: 𝑓(𝑥)=cos2⁡𝑥+2cos⁡𝑥=0=cos⁡𝑥(cos⁡𝑥+2)=0
The term in parentheses is always positive, so has no zeros. (Remember that the most negative cos⁡𝑥 can be is −1.)

Hence the function is zero only when cos⁡𝑥 =0. On the interval [0,2𝜋], this occurs for 𝑥=𝜋2 and 3𝜋2✓

Recall 𝑓(𝑥) =cos2⁡𝑥 +2cos⁡𝑥. Then

𝑓′(𝑥)=2cos⁡𝑥(−sin⁡𝑥)+2(−sin⁡𝑥)=(−2sin⁡𝑥)(cos⁡𝑥+1)

The critical points, such that 𝑓′(𝑐) =0, occur when either term in parentheses is zero on our interval [0,2𝜋]: −2sin⁡𝑐=0𝑐=0,𝜋,2𝜋orcos⁡𝑐+1=0cos⁡𝑐=−1𝑐=𝜋 Our critical points are thus 𝑐 =0, 𝜋, and 2𝜋.

To determine where 𝑓′(𝑥) is positive and where it is negative, we note that ( −2sin⁡𝑥) is negative on the interval (0,𝜋), and positive on the interval (𝜋,2𝜋). The term (cos⁡𝑥 +1) is positive on the interval (0,𝜋), and positive on the interval (𝜋,2𝜋): Hence the product ( −2sin⁡𝑥)(cos⁡𝑥 +1)is negative on the first interval, and positive on the second:

We thus see by the First Derivative Test that the function has a minimum at 𝑥 =𝜋. ✓

From part (b), 𝑓′(𝑥) =( −2sin⁡𝑥)(cos⁡𝑥 +1). Then

𝑓″(𝑥)=(−2cos⁡𝑥)(cos⁡𝑥+1)+(−2sin⁡𝑥)(−sin⁡𝑥)=−2cos2⁡𝑥−2cos⁡𝑥+2sin2⁡𝑥=2[−cos2⁡𝑥−cos⁡𝑥+sin2⁡𝑥]=2[−cos2⁡𝑥−cos⁡𝑥+(1−cos2⁡𝑥)]=2[−2cos2⁡𝑥−cos⁡𝑥+1] Then 𝑓″(𝑥) =0 when:

𝑓″(𝑥)=0=−2cos2⁡𝑥−cos⁡𝑥+1=(−2cos⁡𝑥+1)(cos⁡𝑥+1)

Thus 𝑓″(𝑥) =0 when either term in parentheses is zero: (2cos⁡𝑥−1)=0cos⁡𝑥=12𝑥=𝜋3,5𝜋3or(cos⁡𝑥+1)=0cos⁡𝑥=−1𝑥=𝜋

Our critical points are thus 𝑥 =𝜋3, 𝜋, and 5𝜋3.

Recall from above that 𝑓″(𝑥) =2( −2cos⁡𝑥 +1)(cos⁡𝑥 +1).

To determine where 𝑓′(𝑥) is positive and where it is negative, we note that the term ( −2cos⁡𝑥 +1) is negative on the interval (0,𝜋3) since cos⁡𝑥 >12 there. Similarly, ( −2cos⁡𝑥 +1) is positive on the interval (𝜋3,5𝜋3) since cos⁡𝑥 <12 there. And then on the final interval, (5𝜋3,2𝜋), the term ( −2cos⁡𝑥 +1) is again negative since again cos⁡𝑥 >12 there.

By contrast, the term (cos⁡𝑥 +1) is positive everywhere, except at 𝑥 =𝜋 where it is zero. Hence the product is negative on the intervals (0,𝜋3) and (5𝜋3,2𝜋), and positive on the intervals (𝜋3,𝜋) and (𝜋,5𝜋3): Therefore 𝑓 is concave up on the intervals (𝜋3,𝜋) and (𝜋,5𝜋3). ✓