10 MAXIMUM AND MINIMUM

f(x)  =  x^{2} − 6x + 5 
f(3)  =  3^{2} − 6· 3 + 5 
=  −4. 
The extreme value is −4. To see whether it is a maximum or a minimum, in this case we can simply look at the graph.
f(x) is a parabola, and we can see that the turning point is a minimum.
By finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (3, −4).
But we will not always be able to look at the graph. The algebraic condition for a minimum is that f '(x) changes sign from − to + . We see this at the points E, B, F above. The value of the slope is increasing.
Now to say that the slope is increasing, is to say that, at a critical value, the second derivative (Lesson 9)  which is rate of change of the slope  is positive.
Again, here is f(x):
f(x)  =  x^{2} − 6x + 5. 
f '(x)  =  2x − 6. 
f ''(x)  =  2. 
f '' evaluated at the critical value 3  f ''(3) = 2  is positive. This tells us algebraically that the critical value 3 determines a minimum.
Sufficient conditions
We can now state these sufficient conditions for extreme values of a function at a critical value a:
The function has a minimum value at x = a if f '(a) = 0
and f ''(a) = a positive number.
The function has a maximum value at x = a if f '(a) = 0
and f ''(a) = a negative number.
In the case of the maximum, the slope of the tangent is decreasing  it is going from positive to negative. We can see that at the points C, A, D.
Example 2. Let f(x) = 2x^{3}− 9x^{2} + 12x − 3.
Are there any extreme values? First, are there any critical values  solutions to f '(x) = 0  and do they determine a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum? Where are the turning points?
Solution. f '(x) = 6x^{2} − 18x + 12  =  6(x^{2} − 3x + 2) 
=  6(x − 1)(x − 2)  
=  0 
implies:
x = 1 or x = 2.
Those are the critical values. Does each one determine a maximum or does it determine a minimum? To answer, we must evaluate the second derivative at each value.
f '(x)  =  6x^{2} − 18x + 12. 
f ''(x)  =  12x − 18. 
f ''(1)  =  12 − 18 = −6. 
The second derivative is negative. The function therefore has a maximum at x = 1.
To find the ycoördinate  the extreme value  at that maximum we evaluate f(1):
f(x)  =  2x^{3}− 9x^{2} + 12x − 3 
f(1)  =  2 − 9 + 12 − 3 
=  2. 
The maximum occurs at the point (1, 2).
Next, does x = 2 determine a maximum or a minimum?
f ''(x)  =  12x − 18. 
f ''(2)  =  24 − 18 = 6. 
The second derivative is positive. The function therefore has a minimum at x = 2.
To find the ycoördinate  the extreme value  at that minimum, we evaluate f(2):
f(x)  =  2x^{3} − 9x^{2} + 12x − 3. 
f(2)  =  16 − 36 + 24 − 3 
=  1. 
The minimum occurs at the point (2, 1).
Here in fact is the graph of f(x):
Solutions to f ''(x) = 0 indicate a point of inflection at those solutions, not a maximum or minimum. An example is y = x^{3}. y'' = 6x = 0 implies x = 0. But x = 0 is a point of inflection in the graph of y = x^{3}, not a maximum or minimum.
Another example is y = sin x. The solutions to y'' = 0 are the multiplies of π, which are points of inflection.
Problem 1. Find the coördinates of the vertex of the parabola,
y = x^{2} − 8x + 1.
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
y' = 2x − 8 = 0.
That implies x = 4. That's the xcoördinate of the vertex. To find the ycoördinate, evaluate y at x = 4:
y = 4^{2} − 8· 4 + 1 = −15.
The vertex is at (4, −15).
Problem 2. Examine each function for maxima and minima.
a) y = x^{3} − 3x^{2} + 2.
y' = 3x^{2} − 6x = 3x(x − 2) = 0 implies
x = 0 or x = 2.
y''(x) = 6x − 6.
y''(0) = −6.
The second derivative is negative. That means there is a maximum at x = 0. That maximum value is
y(0) = 2.
Next,
y''(2) = 12 − 6 = 6.
The second derivative is positive. That means there is a minimum at x = 2. That minimum value is
y(2) = 2^{3} − 3· 2^{2} + 2 = 8 − 12 + 2 = −2.
b) y = −2x^{3} − 3x^{2} + 12 x + 10.
At x = 1 there is a maximum of y = 17.
At x = −2 there is a minimum of y = −10.
c) y = 2x^{3} + 3x^{2} + 12 x − 4.
Since f '(x) = 0 has no real solutions, there are no extreme values.
d) y = 3x^{4}− 4x^{3} − 12x^{2} + 2.
At x = 0 there is a maximum of y = 2.
At x = −1 there is a minimum of y = −3.
At x = 2 there is a minimum of y = −30.
Next Lesson: Applications of maximum and minimum values
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