How Do You Know if a Function Has a Maximum or Minimum Value
10
MAXIMUM AND MINIMUM
VALUES
The turning points of a graph
W E SAY THAT A FUNCTION f(x) has a relative maximum value at x = a,
if f(a) is greater than any value immediately preceding or follwing.
We call information technology a "relative" maximum considering other values of the function may in fact be greater.
Nosotros say that a function f(x) has a relative minimum value at x = b,
if f(b) is less than whatever value immediately preceding or follwing.
Once again, other values of the office may in fact be less. With that understanding, then, we volition drop the term relative.
The value of the role, the value of y, at either a maximum or a minimum is chosen an extreme value.
Now, what characterizes the graph at an extreme value?
The tangent to the bend is horizontal. We see this at the points A and B. The gradient of each tangent line -- the derivative when evaluated at a or b -- is 0.
f '(x) = 0.
Moreover, at points immediately to the left of a maximum -- at a pointC -- the slope of the tangent is positive:f '(x) > 0. While at points immediately to the right -- at a point D -- the slope is negative:f '(ten) < 0.
In other words, at a maximum, f '(ten) changes sign from + to − .
At a minimum, f '(x) changes sign from − to + . We can run into that at the points E and F.
Nosotros tin can besides observe that at a maximum, at A, the graph is concave downwards. (Topic 14 of Precalculus.) While at a minimum, at B, it is concave upward.
A value of ten at which the part has either a maximum or a minimum is called a critical value. In the figure --
-- the critical values are x =a and x =b.
The critical values decide turning points, at which the tangent is parallel to the x-axis. The disquisitional values -- if any -- will be the solutions tof '(ten) = 0.
Example 1. Permitf(x) = 10 two − 610 + 5.
Are there whatever disquisitional values -- any turning points? If so, do they make up one's mind a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum?
Solution.f '(x) = 210 − half dozen = 0 impliesx = 3. (Lesson 9 of Algebra.)
x = 3 is the only disquisitional value. It is the x-coördinate of the turning point. To decide the y-coördinate, evaluate f at that critical value -- evaluate f(3):
| f(ten) | = | x ii − 6x + 5 |
| f(three) | = | three2 − 6· 3 + 5 |
| = | −4. | |
The farthermost value is −four. To run across whether it is a maximum or a minimum, in this case we can but await at the graph.
f(x) is a parabola, and we can see that the turning indicate is a minimum.
Past finding the value of x where the derivative is 0, so, nosotros take discovered that the vertex of the parabola is at (three, −4).
Only we will non always be able to await at the graph. The algebraic condition for a minimum is that f '(ten) changes sign from − to + . Nosotros 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 2d derivative (Lesson 9) -- which is rate of change of the slope -- is positive.
Over again, here isf(ten):
| f(x) | = | ten two − vix + 5. |
| f '(x) | = | iiten − half-dozen. |
| f ''(x) | = | 2. |
f '' evaluated at the critical value three -- f''(3) = 2 -- is positive. This tells us algebraically that the critical value 3 determines a minimum.
Sufficient conditions
We tin can at present state these sufficient conditions for extreme values of a function at a disquisitional value a:
The function has a minimum value at 10 =a if f '(a) = 0
and f ''(a) = a positive number.
The part has a maximum value at ten =a if f '(a) = 0
and f ''(a) = a negative number.
In the case of the maximum, the slope of the tangent is decreasing -- information technology is going from positive to negative. We can encounter that at the points C, A, D.
Example ii. Permitf(x) = 2x iii− 910 2 + 12x − 3.
Are there any extreme values? First, are at that place any disquisitional values -- solutions to f '(10) = 0 -- and do they decide 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) = 610 two − eighteenx + 12 | = | 6(10 2 − 3x + ii) |
| = | six(ten − ane)(x − 2) | |
| = | 0 | |
implies:
x = 1 orten = 2.
(Lesson 37 of Algebra.)
Those are the disquisitional 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) | = | 6ten 2 − xviiix + 12. |
| f ''(x) | = | 1210 − 18. |
| f ''(ane) | = | 12 − 18 = −6. |
The second derivative is negative. The part therefore has a maximum at x = 1.
To find the y-coördinate -- the extreme value -- at that maximum we evaluatef(1):
| f(x) | = | twox iii− 9x 2 + 12ten − 3 |
| f(ane) | = | 2 − 9 + 12 − 3 |
| = | two. | |
The maximum occurs at the point (ane, two).
Next, does x = two determine a maximum or a minimum?
| f ''(x) | = | 1210 − eighteen. |
| f ''(2) | = | 24 − 18 = six. |
The second derivative is positive. The function therefore has a minimum at x = 2.
To find the y-coördinate -- the extreme value -- at that minimum, we evaluate f(2):
| f(x) | = | 2x iii − ixten two + 12x − 3. |
| f(2) | = | 16 − 36 + 24 − 3 |
| = | 1. | |
The minimum occurs at the point (2, one).
Here in fact is the graph off(x):
Solutions to f ''(x) = 0 signal a point of inflection at those solutions, not a maximum or minimum. An case is y =x 3.y'' = 610 = 0 implies x = 0. Just x = 0 is a point of inflection in the graph of y =x three, 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.
Trouble 1. Find the coördinates of the vertex of the parabola,
y = x 2 − viiiten + 1.
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Practise the problem yourself commencement!
y' = 2x − 8 = 0.
That implies ten = 4. That's the x-coördinate of the vertex. To find the y-coördinate, evaluate y at x = 4:
y = iv2 − viii· 4 + 1 = −xv.
The vertex is at (4, −15).
Problem 2. Examine each function for maxima and minima.
a) y = x 3 − 310 2 + 2.
y' = iii10 2 − 6x = 3x(ten − ii) = 0 implies
x = 0 or x = ii.
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.
Adjacent,
y''(2) = 12 − six = 6.
The second derivative is positive. That ways there is a minimum at ten = 2. That minimum value is
y(two) = 23 − 3· 22 + 2 = 8 − 12 + two = −2.
b) y = −twox 3 − 310 ii + 12 x + ten.
At x = 1 at that place is a maximum of y = 17.
At x = −2 at that place is a minimum of y = −ten.
c) y = two10 3 + 3x two + 12 x − 4.
Since f '(x) = 0 has no real solutions, in that location are no farthermost values.
d) y = 3x iv− 4x 3 − 12x 2 + 2.
At x = 0 there is a maximum of y = 2.
At x = −1 at that place is a minimum of y = −iii.
At x = two there is a minimum of y = −thirty.
Next Lesson: Applications of maximum and minimum values
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