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 - x^{2} + 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) - 2^{2} + 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 - x^{2} + 3
The given function is the equation of parabola. Replace f(x) by y.
y = -x^{2} + 4 x + 3
Write the above equation of parabola in vertex form.
y = -(x^{2} - 4 x - 3)
y = -[x^{2} - 2(x)(2) + 2^{2} - 2^{2} - 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) = 2x^{3} + 3x^{2} - 36x + 1
Solution :
Find the first derivative of f(x).
f'(x) = 2(3x^{2}) + 3(2x) - 36(1) + 0
= 6x^{2} + 6x - 36
Equate the first derivative to zero, that is f'(x) = 0.
6x^{2} + 6x - 36 = 0
Divide both sides by 6.
x^{2} + 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) = 6x^{2} + 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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