How to Find Maximum and Minimum Values of a Function?

To find the maximum and minimum values of a function we find the derivatives of the given function. If the function f(x) ≤ f(a) for all x ∈ D then f(a) is the maximum value of the function and if f(x) ≥ f(a) for all x ∈ D then f(a) is the minimum value of the function.

Steps to Find Maximum and Minimum Values of Function

Steps to find the maximum and minimum value of the function are added below:

• Step 1: Find the first derivative of the function.

• Step 2: Equate the first derivative of a function to 0.

• Step 3: Find the critical points from the above expression.

• Step 4: Find the second derivative of the function.

• Step 5: If by putting the critical point in the second derivative if the resultant value is negative then, it is the the point of maxima and if the resultant value is positive then, it is point of minima.

• Step 6: By substituting the values of maxima and minima points we get the maximum value and minimum value of the function.

This is further explained with the help of an example added below:

Example: Find the maximum and minimum of function y = x³ - 3x + 11

Solution:

y = x³ - 3x + 11

Find first order derivative. Differentiating y

y’ = (d / dx) [x³ - 3x + 11]
⇒ y’ = (d / dx) x³ - (d / dx) (3x) + (d / dx) 11
⇒ y’ = 3x² - 3 + 0
⇒ y’ = 3x² - 3

Now equate y’ = 0, to find the critical points

y’ = 0
⇒ 3x² - 3 = 0
⇒ 3x² = 3
⇒ x² = 1
⇒ x = 1 or x = -1

Critical points are x = 1 and x = -1

Now we will find second derivative to check the critical point is maxima or minima.

y” = (d / dx) [3x² - 3]
⇒ y” = (d / dx) [3x²] - (d /dx) [3]
⇒ y” = 6x – 0
⇒ y” = 6x

Now we will put the values of x and find whether y” is greater than 0 or less than 0.

At x = 1

y” = 6(1) = 6

Since, y” > 0 x = 1 is the minima of y

At x = -1

y” = 6(-1) = -6

Since, y” < 0 x = -1 is the maxima of y

• Maximum of y at x = -1 is, (-1)³ -3(-1) + 11 = -1 + 3 + 11 = 13
• Minimum of y at x = 1 is, (1)³ -3(1) + 11 = 1 - 3 + 11 = 9

Examples of Finding Maximum and Minimum Value of a Function

Example 1: Find the maximum and minimum value of a function f(x) = x³ + 8x² + 16x + 2.

Solution:

f(x) = x³ + 8x² + 16x + 2

⇒ f'(x) = 3x² + 16x + 16

Then, equate f'(x) = 0, to find critical points.

3x² + 16x + 16 = 0
⇒ 3x² + 12x + 4x + 16 = 0
⇒ 3x (x + 4) + 4 (x + 4) = 0
⇒ (x + 4) (3x + 4) = 0
⇒ x = -4, -4/3

Then, find f''(x) of given function

f''(x) = 6x + 16

Put critical points in f''(x) to find the point of maximum and minimum.

x = -4, f''(x) = -8 and x = -4/3, f''(x) = 8

Since, the value is negative for x = -4 so, it is point of maximum and the value is positive for x = -4/3 so it is point of minimum.

Then, put these values in the f(x) to find the maximum and minimum value of the function.

• Maximum value = f(-4) = - 94
• Minimum value = f(-4/3) = -202 /27 = -7.48

Example 2: Find the maximum and minimum value of a function g(x) = (x³ )/3 - (x² )/2 - 6x +2

Solution:

f(x) = (x³)/3 - (x²)/2 - 6x +2
⇒ f'(x) = x² - x - 6

Then, equate f'(x) = 0, to find critical points.

x² - x - 6 = 0
⇒ x² - 3x + 2x - 6 = 0
⇒ x (x - 3) + 2 (x - 3) = 0
⇒ (x + 2) (x - 3) = 0
⇒ x = -2, 3

Then, find f''(x) of given function

f''(x) = 2x - 1

Put critical points in f''(x) to find the point of maximum and minimum.

x = -2, f''(x) = -5 and x = 3, f''(x) = 5

Since, the value is negative for x = -2 so, it is point of maximum and the value is positive for x = 3 so it is point of minimum.

Then, put these values in the f(x) to find the maximum and minimum value of the function.

• Maximum value = f(-2) = 28/3
• Minimum value = f(3) = -23/2

Read More,

• Calculus
• Differential Calculus
• Maxima or Minima
  • Absolute Minima and Maxima
  • Relative Minima and Maxima