HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A FUNCTION

The following steps would be useful to find the maximum and minimum value of a function using first and second derivatives. 

Step 1 : 

Let f(x) be a function. Find the first derivative of f(x), which is f'(x). 

Step 2 : 

Equate the first derivative f'(x) to zero and solve for x, which are called critical numbers. 

Step 3 : 

Find the second derivative of f(x), which is f"(x). 

Step 4 : 

Substitute the critical numbers found in step 2 in the second derivative f"(x).

Step 5 :

If f"(x) < 0 for some value of x, say x = a, then the function f(x) is maximum at x = a. 

If f"(x) > 0 for some value of x, say x = b, then the function f(x) is minimum at x = b. 

Step 6 : 

To get maximum and minimum values of the function substitute x = a and x = b in f(x). 

Maximum value = f(a)

Minimum value = f(b)

Example 1 :

Determine the maximum value of the function :

f(x) = 4x - x2 + 3

Solution :

Find the first derivative of f(x). 

f'(x) = 4(1) - 2x + 0

= 4 - 2x

Equate the first derivative to zero, that is f'(x) = 0.

4 - 2x = 0

2(2 - x) = 0

2 - x = 0

x = 2

Find the second derivative of f(x).

f'(x) = 4 - 2x

f"(x) = 0 - 2(1)

f"(x) = -2

Substitute the critical number x = 2 in f"(x).

f"(2) = -2 < 0

So, f(x) is maximum at x = 2.

To find the maximum value, substitute x = 2 in f(x).

f(2) = 4(2) - 22 + 3

= 8 - 4 + 3

= 11 - 4

= 7

Therefore the maximum value of the function f(x) is 7.

Justification :

We can justify our answer by graphing the function f(x).

f(x) = 4x - x2 + 3

The given function is the equation of parabola. Replace f(x) by y. 

y = -x2 + 4 x + 3

Write the above equation of parabola in vertex form.

y = -(x2 - 4 x - 3)

y = -[x2 - 2(x)(2) + 22 - 22 - 3]

y = -[(x - 2)2 - 4 - 3]

y = -[(x - 2)2 - 7]

y = -(x - 2)2 + 7

The above equation is in the form y = a(x - h)2 + k.

a = -1

Vertex (h, k) = (2, 7)

Because 'a' is negative the parabola opens down. So, we have only the maximum value for y, that is the y-coordinate at the vertex, which is 7.

The answer is justified.

Example 2 :

Determine the maximum and minimum values of the function :

f(x) = 2x3 + 3x2 - 36x + 1

Solution :

Find the first derivative of f(x). 

f'(x) = 2(3x2) + 3(2x) - 36(1) + 0

= 6x2 + 6x - 36

Equate the first derivative to zero, that is f'(x) = 0.

6x2 + 6x - 36 = 0

Divide both sides by 6.

x2 + x - 6 = 0

Factor and solve. 

(x - 2)(x + 3) = 0

x - 2 = 0

x = 2

x + 3 = 0

x = -3

Find the second derivative of f(x).

f'(x) = 6x2 + 6x - 36

f"(x) = 6(2x) + 6(1) - 0

f"(x) = 12x + 6

Substitute the critical numbers x = 2 and x = -3 in f"(x).

f"(2) = 12(2) + 6

= 24 + 6

= 30 > 0

f"(-3) = 12(-3) + 6

= -36 + 6

= -30 < 0

When x = 2, f"(x) > 0, the function f(x) is minimum at

x = 2 

When x = -3, f"(x) > 0, the function f(x) is maximum at

x = -3

To find the maximum and minimum values of the given function, substitute x = -3 and x = 2 in f(x).

Maximum value : 

f(-3) = 2(-3)3 + 3(-3)2 - 36(-3) + 1

= 2(-27) + 3(9) + 108 + 1

= -54 + 27 + 108 + 1

= 82

Minimum value : 

f(2) = 2(2)3 + 3(2)2 - 36(2) + 1

= 2(8) + 3(4) - 72 + 1

= 16 + 12 - 72 + 1

= -43

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