Solved Examples on Critical Numbers of a Function

Example 1: Find the critical numbers of the function f(x) = x² + 2x.

Solution:

First we will find the first derivative of f(x) i.e., f'(x)

f'(x) = 2x + 2

Now equate f'(x) to zero

2x + 2 = 0

Now solve the above equation and obtain value of x.

2x = -2

x = -1

So, the critical number for the given function f(x) is -1.

Example 2: Find the critical numbers of the function f(x) = (x³ / 3) – 2x² + 4x + 1.

Solution:

First we will find the first derivative of f(x) i.e., f'(x)

f'(x) = 3(x² / 3 ) – 4x + 4

f'(x) = x² – 4x + 4

Now equate f'(x) = 0

x² – 4x + 4 = 0

(x – 2)² = 0

x = 2

So, the critical number of the function f(x) is 2.

Example 3: Find the critical numbers of function p(x) = x ln x.

Solution:

First we will find the first derivative of p(x) i.e., p'(x)

To differentiate p(x), we will use the product rule, which states that if u(x) and v(x) are functions of x, then (uv)′=u′v + uv′

Here, u(x) = x and v(x) = ln⁡x

u′(x) = 1 and v'(x) = 1/x

Using the product rule we have:

p′(x) = (x)′ln⁡x + x(ln⁡x)

p′(x) = 1 ⋅ ln⁡x + x ⋅ 1/x

p'(x) = ln x + 1

Now equate p'(x) = 0

1 + ln x = 0

ln x = -1

x = e^-1

The critical number of p(x) is e^-1 .

Example 4: Determine the critical number of the function a(x) = sin 2x.

Solution:

First, we will find a'(x)

a'(x) = 2 cos 2x

Now put a'(x) = 0

2 cos 2x = 0

cos 2x = 0

2x = π / 2

x = π / 4

Example 5: Find the critical number of f(x) = 2 sin x – x

Solution:

First, find f'(x)

f'(x) = 2cos x – 1

Now put f'(x) = 0

2 cos x – 1 = 0

2 cos x = 1

cos x = 1 / 2

x = π / 3

Example 6: Find the critical number of function f(x) = x³ – 3x² + 2x

Solution:

First find f'(x)

f'(x) = 3x² – 6x + 2

Now put f'(x) = 0

3x² – 6x + 2 = 0

Solve this quadratic equation:

[Tex]x = \frac{6\pm\sqrt{(-6)^2 – 4\cdot3\cdot2)}}{2\cdot3}[/Tex]

[Tex]x = \frac{6 \pm\sqrt{36-24}}{6}[/Tex]

[Tex]x = \frac{6 \pm\sqrt{12}}{6}[/Tex]

[Tex]x = \frac{6 \pm2\sqrt3}{6}= 1 \pm \frac{\sqrt{3}}{3}[/Tex]

So, the critical numbers are:

[Tex]x = 1 + \frac{\sqrt 3}{3}\ and \ 1 – \frac{\sqrt 3}{3} [/Tex]

Example 7: Find the critical numbers of function f(x) = (x/ 2) + (2 / x)

Solution:

f(x) = (x / 2) + (2 / x)

First find f'(x)

f'(x) = 1/2 + 2 (-1/x²)

f'(x) = 1/2 – 2/x²

Now put f'(x) = 0

1/2 – 2/x² = 0

1/2 = 2/x²

x² = 4

x = 2, -2

Example 8: Find the critical numbers of function f(x) = tan x – 2x

Solution:

First find f'(x)

f'(x) = sec² x – 2

Now, put f'(x) = 0

sec² x – 2 = 0

sec² x = 2

Taking root both sides

sec x = √2

x = π / 4 or 3π / 4

Example 9: Find the critical numbers of function f(x) = sin x + cos x

Solution:

f(x) = sin x + cos x

First find f'(x)

f'(x) = cos x – sin x

Now put f'(x) = 0

cos x – sin x = 0

Divide both sides by cos⁡x:

1 = tan x

x = π / 4 + kπ where k is any integer.

Example 10: Find the critical numbers of function f(x) = sin x + (1/2) cos x

Solution:

First find f'(x)

f'(x) = cos x – (1/2)sin x

Now, put f'(x) = 0

cos x – (1/2)sin 2x = 0

cos x = (1/2) sin x

Divide both sides by cos⁡x:

1 = (1/2)tan x

tanx = 2

The general solution for tan⁡x = 2 is:

x = tan⁡⁻¹(2) + kπ, where k is any integer.

Read More,

• How to Find Maximum and Minimum Values of a Function
• How to find the Equation of a Quadratic Function
• How to Find Rank of a 3×3 Matrix

To find the critical value of a function we simply take its derivative, set it to zero, and solve for x. The values of x for which the derivative is zero are critical numbers.

In this article, we will cover critical numbers and how to find critical numbers of function. We will also solve some problems related to how to find critical numbers of a function.