How derivative affects the shape of a curve

 

HOW DERIVATIVES AFFECT THE SHAPE OF A CURVE

We know that f '(x) represents the slope of the curve y = f(x) at the point (x, f(x)). It tells us the direction in which the curve proceeds at each point. It is reasonable to expect that information about f '(x) will provide us with information about f(x). The derivative of ƒ can tell us where a function is increasing or decreasing.

Notice that:

• Between A and B and between C and D, the tangent lines have positive slope and so f '(x) > 0.

• Between B and C the tangent lines have negative slope and so f '(x) < 0

• Thus it appears that f increases when f '(x) is positive and decreases when f '(x) is negative.

Increasing/Decreasing test.

• If f '(x) > 0 on an interval I, then f is increasing function on that interval I.

• If f '(x) < 0 on an interval I, then f is decreasing function on that interval I.

 

First derivative test

Suppose that c is a critical point of a function.

• If f ' changes from +ive to ‒ive at c, then ƒ has the local maximum at c.

• If f ' changes from ‒ive to +ive at c, then ƒ has the local minimum at c.

• If f ' does not change sign at c, then f has no local maximum or minimum at c.

The first derivate test is a consequence of the increasing/decreasing test

Definition

If the graph of f, lies above all of its tangents on an interval I, then the curve is called concave upward on I

If the graph of f, lies below all of its tangents on an interval I, then the curve is called concave downward on I

If a function ƒ has a local extremum at a point c, then c must be a critical point of f. However, a function is not guaranteed to have a local extremum at a critical point. For example, f(x) = x³ has a critical point at x=0 since f '(x) = 3x² is zero at x = 0, but ƒ does not have a local extremum at x=0. Using the results from the previous section, we are now able to determine whether a critical point of a function actually corresponds to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a graph by describing whether the graph of a function curves upward or curves downward.

 

Concavity test

• If ƒ"(x) > 0 for all in I, then the graph of ƒ is concave upward on I [f '(x) is increaing]

• If f"(x) < 0 for all in I, then the graph of ƒ is concave downward on I [f '(x) is decreaing]

The below figure shows the graphs of two increasing functions on (a, b). Both graphs join point A to point B but they look different because they bend in different directions. How can we distinguish between these two types of behavior?. In the first figure the curve lies above the tangents and is called concave upward on (a, b). In the second figure the curve lies below the tangents and is called concave downward on (a, b).

Definition

A point P on a curve y = f(x) is called an inflection point if f is continuous and the curve changes from concave upward to concave downward or from concave downward to concave upward at P.

If f "(c) = 0, then f(x) has a point of inflection at x = c. The point of inflection is (c, f(c)).

Note:

• If a curve has a tangent at a point of inflection, then the curve crosses its tangent there.

• There is a point of inflection at any point where the second derivative changes sign

 

Second derivative test

Suppose f " be a continuous near c.

• If f '(c) = 0 and f"(c) < 0, then f(x) has local maximum at x = c

• If f '(c) = 0 and f"(c) > 0, then f(x) has local minimum at x = c

 

 

Example 138. Find the function f(x)=3x⁴‒4x³‒12x² +5 is increasing and where it is decreasing.

Solution:

ƒ'(x) = 12x³ – 12x² – 24x = 12x(x² − x − 2) = 12x (x − 2)(x + 1)

To find the critical point:

 f '(x) = 0

12x (x‒2)(x + 1) = 0

 x = ‒1,0,2 are critical points

We divide the real line into intervals whose end points are the critical points x = ‒1,0,2 and complete the following table.

The graph of ƒ shown in figure confirms the information in the chart

 

Example 139. Find the local maximum and minimum values of the function g(x) = x+2sinx, 0 ≤ x ≤ 2π

Solution: To find the critical numbers of g, we differentiate

 g'(x) = 1 + 2cosx

So g'(x) = 0 where cosx = ‒ 1/2

The solutions of this equation are 2π/3 and 4π/3.

We divide the real line into intervals whose end points are the critical points x = 2π/3,4π/3 and complete the following table.

Because g'(x) changes from positive to negative at 2π/3, the First Derivative Test tells us that there is a local maximum at 2π/3 and the local maximum value is

Likewise, g'(x) changes from negative to positive at 4π/3 and so

is a local minimum value. The graph of g in figure supports our conclusion.

 

Example 140. Find where the function f(x) = 2x³ +3x²‒36x

a) the intervals on which increasing/decreasing

b) local maximum and minimum values of f

c) intervals of concavity and inflection point.

Solution: (a) The given function

f(x)= 2x³ + 3x²‒36x and

 f '(x) = 6x²+6x‒36

 =6(x²+x‒6) = 6(x+3)(x − 2)

 f "(x) = 12x+6 =6(2x+1)

To find the critical point:

 f '(x) = 0

6(x+3)(x‒2)= 0

x = ‒3,2 are critical points

We divide the real line into intervals whose end points are the critical points x = ‒3,2 and complete the following table.

(b) Now we apply the first derivative test to find the local extreme values f(x) changes from increasing to decreasing at x = ‒3. Thus the function has a local maximum at x = ‒3 and the local maximum value is

 ƒ(−3) = 2(‒3)³ + 3(−3)² ‒ 36(−3) = 81

f(x) changes from decreasing to increasing at x = 2. Thus the function has a local minimum at x = 2 and the local minimum value is

 ƒ(2) = 2(2)³ + 3(2)² ‒ 36(2) = ‒44

(c) To find point of inflection:

 f '(x) = 0

6(2x + 1) = 0

2x+1=0

⇒ 2x = ‒1

 x = ‒ 1/2

We divide the real line into intervals whose end points are the critical point x = ‒ ½ and complete the following table.

Since the curve changes from concave downward to concave upward at x = ‒ ½ the points of inflection is ( ‒1/2 , f(‒1) )

ƒ(−1/2) = 2 ( − 1/2 )² + 3 (− 1/2 )² ‒ 36 (− 1/2) = 37/2

Hence the point of inflection is (‒1/2 , 37/2).

 

Example 141. f(x) = e^2x + e^‒x

a) Find the intervals on which increasing/decreasing

b) Local maximum and minimum values of ƒ

c) intervals of concavity and inflection point

Solution:

(a) The given function is

f(x) = e^2x+e^‒x and

ƒ'(x) = 2e^2x ‒ e^‒x

ƒ"(x) = 4e^2x + e^‒x

To find the critical point:

 f '(x) = 0

2e^2x ‒ e^‒x = 0

2e^2x = e^‒x

2 = e^‒x / e^2x

⇒ 2 = e^‒3x

log2 = loge^‒3x = ‒3x

‒3x= log 2

 x = ‒1/3 log2

We divide the real line into intervals whose end points are the critical point x = ‒1/3 log 2 and complete the following table.

(b) Now we apply the first derivative test to find the local extreme values f(x) changes from decreasing to increasing at x = ‒ 1/3 log 2. Thus the function has a local minimum at x= ‒ 1/3 log 2 and the local minimum value is

 f(‒1/3 log 2) = e^{‒2/3log 2} + e^{1/3log 2}

= e^{log 2 pw(‒2/3)} + e^{log 2 pw(1/3)}

= 2^‒2/3+ 2^1/3

= 1.889

(c) To find point of inflection:

 f "(x) = 0

 f "(x) = 4e^2x + e^‒x

Since 4e^2x > 0 and e^‒x > 0 for all values of x in R, f ''(x) > 0.Hence f is concave upward on (‒∞, ∞). There is no point of an inflection.

 

Example 142. f(x) = cos²x ‒ 2sinx

a) Find the intervals on which increasing/decreasing

b) Local maximum and minimum values of f

c) intervals of concavity and inflection point

Solution: a) The given function is

f(x) = cos²x ‒ 2 sin x

and

 f '(x) = 2 cos x(‒ sin x) ‒ 2 cos x

= ‒2 cosx sinx ‒ 2 cosx

=‒2 cosx (1 + sinx)

To find the critical point:

 f '(x) = 0

 ‒2 cosx(1 + sin x) = 0

cos x = 0 and 1 + sin x = 0

cos x = 0

⇒ x = π/2, 3π/2

1 + sin x = 0

sin x = ‒1

⇒ x= 3π / 2

Therefore the critical points are π/2, 3π/2

Each of this critical points lies in the interval [0, 2π].

We divide the real line into intervals whose end points are the critical points x = π/2, 3π/2 and complete the following table.

(b) Now we apply the first derivative test to find the local extreme values f(x) changes from decreasing to increasing at x = π/2. Thus the function has a local minimum at x= π/2 and the local minimum value is

 f(π/2) = cos²(π/2) ‒ 2sin(π/2)

= 0‒2(1) = ‒2

f(x) changes from increasing to decreasing at x = 3π/2. Thus the function has a local maximum at x = 3π/2 and the local maximum value is

f(3π/2) = cos²(3π/2) ‒ 2 sin(3π/2)

= 0‒2(‒1) = ‒2

c) To find point of inflection

f ''(x) = −2 cosx (cos x) + (sin x + 1) 2sinx

= ‒2 cos²x + 2 sin²x + 2 sinx

= 2(‒cos²x + sin²x + sinx)

= 2(‒(1 − sin²x) + sin²x + sinx)

= 2(‒1 + sin²x + sin²x + sinx)

= 2(2 sin²x + sinx ‒ 1)

= 2(2sinx – 1)(sinx +1)

 ƒ"(x) = 0

2(2sinx – 1)(sinx+1)=0

2sinx ‒ 1 = 0 and sinx + 1=0

2sinx – 1 = 0

2sinx – 1 = 0

2sinx = 1

sin x = ½

 x = π/6, π ‒ π/6

x = π/6, [6π ‒ π]/6

x = π/6, 5π/6

Also sinx + 1 = 0

sin x = ‒1

 x = 3π/2

 x = π/6, 5π/6, 3π/2

We divide the real line into intervals whose end points are the critical points x = π/6, 5π/6, 3π/2 and complete the following table.

Since the curve changes from concave downward to concave upward at x = π/6 and the curve changes from concave upward to concave downward at x = 5π/6

Thus, the points of inflection are (π/6, ,f(π/6)) and (5π/6, ,f(5π/6)).

Since,

 f(π/6) = cos²(π/6) ‒ 2sin(π/6)

= (√3/2)² ‒ 2(1/2)

= 3/4 ‒ 1

= ‒ 1/4

f(5π/6) = cos²(5π/6) ‒ 2sin(5π/6)

= (√3/2)² ‒ 2(1/2)

= 3/4 ‒ 1

= ‒ 1/4

Hence, the points of inflection are (π/6,‒1/4) and (5π/6, ‒1/4)

 

Example 143. A particle is moving along a horizontal coordinate line (positive to the right) with position function s(t) = 2t³ ‒ 14t² + 22t ‒ 5, t≥0, Find the velocity and acceleration, and describe the motion of the particle.

Solution:

The displacement is

s(t) = 2t³‒14t² + 22t ‒ 5

When the function s(t) is increasing, the particle is moving to the right; when s(t) is decreasing, the particle is moving to the left.

To find the critical points of s, we differentiate

The velocity is

 v(t) = s'(t) = 6t² ‒ 28t + 22

= 2(3t² ‒ 14t + 11) = 2(t − 1)(3t ‒ 11)

So s'(t) = 0

2(t ‒ 1)(3t ‒ 11) = 0

t = 1 or t = 11/3.

We divide the real line into intervals whose end points are the critical points x =

1,11/3 and complete the following table.

The particle is moving to the right in the time intervals [0, 1) and (11/3,∞) and moving to the left in (1, 11/3).

The acceleration is

a(t) = v'(t) = s"(t) = 12t ‒ 28 = 4(3t ‒ 7)

momentarily stationary (at rest) at t = 1 and t = 11/3. The acceleration a(t) = s"(t) = 4(3t ‒ 7) is zero when t = 7/3.

The particle starts out moving to the right while slowing down, and then reverses and begins moving to the left at t = 1 under the influence of the leftward acceleration over the time interval [0,7/3). The acceleration then changes direction at t = 7/3 but the particle continues moving leftward, while slowing down under the rightward acceleration. At t = 11/3 the particle reverses direction again: moving to the right in the same direction as the acceleration, so it is speeding up.

 

Example 144. Sketch a graph of the function

 f(x) = x⁴ − 4x³ + 10

using the following steps.

(a) Identify where the extrema of ƒ occur.

(b) Find the intervals on which ƒ is increasing and the intervals on which f is decreasing.

(c) Find where the graph of ƒ is concave up and where it is concave down.

(d) Sketch the general shape of the graph for f.

(e) Plot some specific points, such as local maximum and minimum points, points of inflection, and intercepts. Then sketch the curve.

Solution:

The function f  is continuous since f '(x) = 4x³‒ 12x² exists. The domain of ƒ is (‒∞, ∞), and the domain of f ' is also (‒∞, ∞), Thus, the critical points of f occur only at the zeros of f '. Since

 f '(x) = 4x³ ‒ 12x² = 4x²(x − 3).

To find the critical point:

 f '(x) = 0

4x²(x‒3)=0

 x = 0,3 are critical points

We divide the real line into intervals whose end points are the critical points x = 0,3 and complete the following table

a) Using the First Derivative Test for local extrema and the table above, we see that there is no extremum at x = 0 and a local minimum at x = 3.

(b) Using the table above, we see that f is decreasing on (‒∞, 0] and [0,3], and increasing on [3,∞).

(c) f ''(x) = 12x² ‒ 24x = 12x(x‒2)

To find the point of inflection:

ƒ"(x) = 0

12x(x‒2)=0

x = 0 or x = 2.

We divide the real line into intervals whose end points are the critical points x = 0,2 and complete the following table

We see that ƒ is concave up on the intervals (‒∞, 0) and (2,∞), and concave down on (0,2).

(d) Summarizing the information in the last two tables, we obtain the following.

The general shape of the graph

e) The graph of f(x) is shown below:

 

Example 145. Discuss the curve y=x⁴‒4x³ with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve.

Solution:

Let f(x) = x⁴ ‒ 4x³. Then

 f '(x) = 4x³‒12x² = 4x²(x − 3)

ƒ"(x) = 12x² ‒ 24x = 12x(x − 2).

The critical numbers are given by

 f '(x) = 0

4x²(x‒3)=0

x = 0 and x = 3.

We divide the real line into intervals whose end point are the critical points x=0,3 and complete the following table.

Now we apply the first derivative test to find the local extreme values.

 f '(x) changes from decreasing to increasing at x = 3. Thus the function has a local minimum at x = 3 and the local minimum value is

ƒ(3) = (3)⁴ ‒ 4(3)³

= 81‒108 = ‒27

Since f(x) has no change at x=0. Thus, f(x) has neither local maximum nor local minimum.

For concavity,

ƒ"(x) = 0

12x(x‒2)=0

x = 0,2

We divide the real line into intervals with 0,2 as endpoints and complete the following table.

Since the curve changes from concave upward to concave downward at x = 0 and the curve changes from concave downward to concave upward x=2.

Thus, the points of inflection are (0, f(0)) and (2, f(2)).

Since,

f(0) = (0)⁴‒4(0)³ = 0

ƒ(2) = (2)⁴ − 4(2)³ = 16 – 32 = −16.

Hence the points of inflection are (0,0) and (2,‒16).

The graph of f(x) is shown below:

 

Example 146. Sketch the graph of the function f(x) = x^2/3 (6‒x)^1/3

Solution: The function is

ƒ(x) = x^2/3(6 − x)^1/3

Taking log on both sides and differentiate with respect to x

log f(x) = log(x^2/3(6‒x)^1/3]

= log x^2/3 + log(6‒x)^1/3

= 2/3 logx + 1/3 log(6‒x)

Since f '(x) = 0 when x = 4 and does not exist when x = 0

or x = 6, the critical numbers are 0,4 and 6.

We divide the real line into intervals with 0,2 as endpoints and complete the following table.

We apply the first derivative test to find the local extreme values.

f(x) change from decreasing to increasing at x = 0,Thus, the function has a local minimum x = 0 and local minimum value is

ƒ(0) = (0)^2/3(6‒0)^1/3 = 0

f(x) change from increasing to decreasing at x = 4,Thus, the function has a local maximum x = 4 and local maximum value is

f(4) = (4)^2/3(6‒4)^1/3 = 2^5/3

Since f '(x) has no change at x = 6. Thus, f(x) has neither local minimum nor local maximum.

For concavity, since f "(x) does not exist at x = 0 and x =6 .So the critical points are x = 0,6.

We divide the real line into intervals with 0,6 as endpoints and complete the following table.

Since the curve changes from concave downward to concave upward at x = 6. Thus, the points of inflection are (6,ƒ(6)).

Since, f(6) = (6)^2/3(6 – 6)^1/3 = 0

Hence, the points of inflection is (6,0).

The graph is sketched in following figure. Note that the curve has vertical tangents at (0,0) and (6, 0) because |ƒ'(x)| → ∞ as x → 0 and as x → 6.

 

Example 147. Find the local maximum and minimum values of f(x) = √x‒⁴√x using both the first and second derivative tests.

Solution:

 f(x) = √x ‒ ⁴√x

 f '(x) = 1/2x^‒1/2 ‒ 1/4x^‒3/4

=  [ 2 ⁴√x ‒ 1 ] / [ 4 ⁴√x³ ]

f '(x) = 0

⇒ 2⁴√x −1 = 0

⇒x = 1 / 16

First derivative test:

 f '(x) > 0 when x > 1/16

f '(x) < 0 ⇒.0 < x > 1/16

Since f '(x) changes from negative to positive at x = 1/16 f(x) has local minimum value at x = 1/16.

The local minimum value is

 f(1/16) = ¼ ‒ ½  = ‒ ¼

Second derivative test:

 = ‒16+24=8 > 0

f(x) has local minimum value at x= 1 / 16

f(1/16) = ¼ ‒ ½  = ‒ ¼ = is the local minimum value of f (x).

 

 

 

EXERCISE

 

38. Find the local maximum and minimum values of the function

f(x) = x + 2sinx, 0≤ x ≤ 2π.

Ans: local maximum = 3.83, local minimum = 2.46

 

39. Suppose the derivative of the function f is

 f '(x) = (x+1)²(x‒3)⁵(x‒6)⁴ on what interval is f increasing?

Ans: (3,∞)

 

40. Show that the inflection points of the curve y=x sin x lie on the curve y²(x²+4)=4x².

 

41. If the function f(x) = x³ + ax² + bx has the local minimum value ‒ 2√3/9 at x=1/√3 what are the values of a and b ?

Ans: a = 0, b = ‒1

 

42. Which of the tangent lines to the curve f(x) = x³ + ax² + bx has the smallest slope?

 

43. Use the first and second derivatives of f(x) = e^1/x, together with asymptotes to sketch its graph.