Applications of Differentiation – Curve Sketching Using Algebra and Calculus (First and Second Derivative Tests)

Typically, if you want to know what the graph of a function will look like, you would use a graphing calculator or software.

In case you do not have a graphing calculator, you could create a x-y table, plot the points, and trace the graph that way.

However, sometimes you will not have access to a graphing calculator and creating a x-y table and plotting points will be too time-consuming.

Therefore, it is best to sketch the curve, or draw an approximation of the appearance of the graph from certain clues.

From precalculus, when we sketched the graph of the function, we used concepts such as domain and range, intercepts, and asymptotes. However, to make our sketches more accurate, we will include the first and second derivative tests to determine the intervals of increasing and decreasing, local extrema, intervals of concavity, and inflection points.

In total, to sketch the curve of a graph, we will discuss the following concepts:

  • Domain and range
  • Intercepts
  • Asymptotes
  • Intervals of increasing or decreasing
  • Local maxima and minima
  • Concavity
  • Inflection points

Example 1:

Sketch the function

Solution:

  • Domain and Range

As this is a rational function so, domain of this function F (x) will be:

x2 = 0

x = 0

So, the domain of this function will be (-∞ to 0) U (0 to ∞).

On that same note, x = 0 is a vertical asymptote.

To find the range of rational functions, we have to isolate the term with the highest degree for both the numerator and denominator:

y = x-1/x2

y = x/x2

Since the highest degree on the denominator is bigger than the highest degree in the numerator, the horizontal asymptote will be y = 0.

  • Intercepts:

To find the x-intercept, we have to set y equal to zero and solve for x:

Perform cross multiplication:

0*x2= x – 1

0 = x-1

Solve for x by adding 1 to both sides:

0 + 1 = x + 1 – 1

1 = x

Now we know that the x intercept is x = 1. Plot the point (1,0) on the coordinate plane.

To find the y intercept, set x equal to zero and solve for y:

By solving this, we get the following:

y = -1/0

y = undefined

This means that there is no y intercept of the function, which means that the graph does not intersect the y axis at all.

This further proves that there is a vertical asymptote at x = 0

  • Asymptotes:

When we solved for the domain and range of the function, we discovered that x and y cannot be equal to zero. Therefore,

x = 0 (which is also the y axis) is a vertical asymptote.

y = 0 (which is also the x axis) is a horizontal asymptote.

  • Intervals of increasing/decreasing and local maxima or minima:

Solve for x by first performing cross multiplication:

0 * x3= 2 – x

0 = 2 – x

Add the x to the other side:

x = 2 (first critical number)

Now we need to see what values that will make y’ undefined. We see that in the denominator there is a x3. For rational functions, we know that the denominator cannot be zero. Therefore, to find the x value that makes y’ undefined, set x3= 0

x3= 0

x = 0 (second critical number)

Plot the critical points on the number line:

To find intervals of increasing and decreasing, we need to substitute values inside y’ from our number line.

Let’s try a value in between – ∞ and 0, such as -1:

Since the result is negative, the graph of F(x) will decrease from 2 to ∞. Therefore,

Interval of increase: (0, 2)

Intervals of decrease: (-∞, 0) and (2, ∞)

To find the points local maximum and local minimum, we need to plug in the critical numbers into the original function F(x) to obtain the y coordinates. Let’s start with the critical point x = 0:

Since F(0) is undefined, there is no y coordinate for the critical point x = 0. Therefore, there is no point at x = 0. We must recall that x = 0 is the vertical asymptote. Now we must find the y coordinate of the critical point x = 2:

Since the function F(x) is increasing on the interval of 0 to 2 and decreasing from 2 to ∞, the point (2, ¼) is a local maximum.

  • Intervals of concavity and inflection points:

To find the concavity and inflection point of any function, we have to take the second derivative of the function:

So,

Perform cross multiplication:

0 * x4= 2(x – 3)

0 = 2(x-3)

Divide 2 on both sides:

0/2 = x – 3

0 = x – 3

3 = x

Now we need to find the x values that make y’’ undefined. Since y’’ is a rational function, we need to set the denominator equal to zero:

x4= 0

x = 0

Plot the inflection numbers on the number line:

Since the result is positive, the concavity will be positive from 3 to ∞. Therefore,

Intervals of positive concavity (concave up): (3, ∞)

Intervals of negative concavity (concave down): (-∞, 0) and (0, 3)

To get our first inflection point, we put x = 3 in the real function F (x),

We will get,

From this calculation, our only inflection point is (3, 2/9).

For our second inflection point, plug in x = 0 into the original function F(x), which will give us an undefined answer. Therefore, x = 0 does not have a y coordinate and therefore there is no inflection point at x = 0 (recall once again that x = 0 is a vertical asymptote).

Now that we have found all the clues, we can finally sketch the graph. Let us summarize what we have discovered:

  • x = 1 is the x intercept. The point (1,0) exists on the graph.
  • The domain is (-∞ to 0) U (0 to ∞). x = 0 (the y axis) is the vertical asymptote.
  • y = 0 (the x axis) is the horizontal asymptote.
  • Interval on increasing: (0, 2)
  • Intervals of decreasing: (-∞, 0) and (2, ∞)
  • Local maximum at (2, ¼)
  • Interval of positive concavity (concave up): (3, ∞)
  • Interval of negative concavity (concave down): (-∞, 0) and (0, 3)
  • Inflection point: (3, 2/9)

Use the clues to sketch your interpretation of the graph to the best of your ability. Compare your sketch to the actual graph of F(x):

Example 2:

Sketch the function F(x) = -x3/3+x2

Solution:

  • Domain and Range

We can rewrite this function as y = -1/3x3+x2. As such, this is a polynomial function. Therefore, the domain is (-∞, ∞), which is the entire set of x values. The range is also (-∞, ∞), or the entire set of all y values. Essentially, there is no limit on the domain and range.

Note: For polynomial functions, the domain is the entire set of x values, and the range is the entire yet of y values. There is no restriction on the domain and range.

  • Intercepts:

To find the x-intercept, we have to set y equal to zero and solve for x:

The y intercept is y = 0. Therefore, we know that the point (0,0) exists on the graph.

  • Asymptotes:

Going back to the domain and range of the function, because there is no limit on the domain and range of the function, there will not be any asymptotes for our function.

  • Intervals of increasing/decreasing and local maxima or minima:

As we are given the function:

After taking the first derivative, we have to find the x value that will make y’ equal to zero or undefined:

y’ = -x2+2x

To find the x values that will make y’ equal to zero, set y’ = 0 and solve for x:

0 = -x2+2x

Notice that each term has an x factor. Therefore, we will factor an x from each term:

0 = -x2+2x

0 = (x)(-x+2)

We will set each factor equal to zero:

0 = x (our first critical number is 0)

0 = -x+2

Add the x to both sides:

x = 2 (our second critical number is 2)

Now we need to see what values that will make y’ undefined. Because y’ is a polynomial, there is no x value that will make it undefined (there is no restriction on the domain). Therefore, our critical numbers are 0 and 2.

Plot the critical points on the number line:

To find intervals of increasing and decreasing, we need to substitute values inside y’ from our number line.

Let’s try a value in between – ∞ and 0, such as -1:

y’(-1) = -(-1)2+2(-1)= -1-2= -3

Since the result is negative, the graph of F(x) will decrease from -∞to 0.

Let’s try a number between 0 and 2, such as 1:
y’(1) = -(1)2+2(1)= -1+2=1

Since the result is positive, the graph of F(x) will increase from 0 to 2.

Now let’s try a number between 2 and ∞, such as 3:

y’(3) = -(3)2+2(3) = -9 + 6 = -3

Since the result is negative, the graph of F(x) will decrease from 2 to ∞.

Our second critical point is (2, 1.33)

Since the function F(x) is increasing on the interval of 0 to 2 and decreasing from 2 to , the point (2, 1.33) is a local maximum.

Since the graph is decreasing from (-∞, 0), the point (0,0) is a local minimum).

  • Intervals of concavity and inflection points:

To find the concavity and inflection point of any function, we have to take the second derivative of the function F(x)

Our first derivative is already solved:

F’ (x) = y’ = -x2+2x

Now take the second derivative using the power rule:

y’’ = -(2)x2-1+2x1-1

y’’ = -(2)x1+2x0

y’’ = -2x+2(1)

y’’ = -2x+2

To find the inflection numbers, we need to find the x values that makes y’’ zero or undefined:

Let’s set y’’ equal to zero:

0 = -2x+2

Add 2x on both sides:

-2x = 2

Divide -2 on both sides:

-2x/-2=2/-2

x = -1

Our only inflection number is x = -1.

Now we need to find the x values that make y’’ undefined. Since y’’ is a polynomial, there will not be an x value to make y’’ undefined (there is no restriction on the domain).

Plot the inflection number on the number line:

To find intervals of concavity, we need to substitute values inside y’’ from our number line.

Let’s try a value in between – ∞ and -1, such as -2:

y’’(-2) = -2(-2)+2= 4 +2 = 6

Since the result is positive, the concavity will be positive from – ∞ to -1.

Now, let’s plug in a number between -1 and -∞, such as 2.

y’’(0) = -2(2)+2= -4 +2 = -2

Since the result is negative, the concavity will also be negative from -1 to ∞.

Therefore,

Intervals of positive concavity (concave up): (-∞, 1)

Intervals of negative concavity (concave down): (-1,∞)

To get our inflection point, we put x = -1 in the original function F (x),

Our only inflection point is (-1, 1.33) .

Now that we have found all the clues, we can finally sketch the graph. Let us summarize what we have discovered:

  • The domain is (-∞, ∞)
  • The range is (-∞, ∞)
  • The x intercept is x = 3. The point (3,0) exists on the graph
  • The y intercept is y = 0. The point (0,0) exists on the graph
  • Interval of increasing: (0,2)
  • Intervals of decreasing: (-∞, 0) and (2, ∞)
  • Critical points: (0,0) and (2, 1.33)
  • Intervals of positive concavity: (-∞, 1)
  • Intervals of negative concavity: (-1,∞)
  • Inflection point: (-1, 1.33)

Compare your sketch to the actual graph of F(x):

Summary of Section

Curve Sketching incorporates techniques from both algebra and calculus that can be used to produce a rough idea of the overall shape of a graph given the equation.

The following can be found when sketching a graph:

  • Domain and range
  • Intercepts
  • Asymptotes
  • Intervals of increasing or decreasing
  • Local maxima and minima
  • Concavity
  • Inflection points

Of course, the sketch doesn’t have to be an exact replica of the actual curve, but using algebraic and calculus techniques will help to get as close as possible to the real shape of the curve.

References:
http://www.opentextbookstore.com/appcalc/Chapter2-8.pdf
https://www.math24.net/curve-sketching/
https://magoosh.com/hs/ap-calculus/2017/ap-calculus-10-step-guide-curve-sketching/
https://www.intmath.com/applications-differentiation/5-curve-sketching-differentiation.php

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