how do you find the maximum or minimum value of a function
10
MAXIMUM AND MINIMUM
VALUES
The turning points of a graph
W Due east 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 it a "relative" maximum because other values of the role may in fact be greater.
We say that a role f(x) has a relative minimum value at x = b,
if f(b) is less than whatever value immediately preceding or follwing.
Again, other values of the function may in fact be less. With that understanding, then, we volition drop the term relative.
The value of the function, the value of y, at either a maximum or a minimum is called an farthermost value.
Now, what characterizes the graph at an extreme value?
The tangent to the curve is horizontal. We run into this at the points A and B. The slope 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 '(ten) > 0. While at points immediately to the right -- at a point D -- the slope is negative:f '(x) < 0.
In other words, at a maximum, f '(ten) changes sign from + to − .
At a minimum, f '(x) changes sign from − to + . We can see that at the points E and F.
Nosotros tin also observe that at a maximum, at A, the graph is concave down. (Topic xiv of Precalculus.) While at a minimum, at B, it is concave upward.
A value of x at which the function has either a maximum or a minimum is chosen a disquisitional value. In the figure --
-- the critical values are x =a and ten =b.
The critical values determine turning points, at which the tangent is parallel to the x-axis. The critical values -- if whatever -- will be the solutions tof '(x) = 0.
Example 1. Letf(x) = x 2 − half dozenx + five.
Are there any critical values -- whatsoever turning points? If so, do they determine a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum?
Solution.f '(x) = ii10 − 6 = 0 implies10 = 3. (Lesson 9 of Algebra.)
x = 3 is the only critical value. Information technology is the 10-coördinate of the turning point. To determine the y-coördinate, evaluate f at that critical value -- evaluate f(3):
| f(x) | = | x two − vix + five |
| f(3) | = | threetwo − half-dozen· 3 + 5 |
| = | −iv. | |
The extreme value is −iv. To see whether information technology is a maximum or a minimum, in this example we can simply look at the graph.
f(10) 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 (three, −4).
But we will not always be able to look at the graph. The algebraic status for a minimum is that f '(10) changes sign from − to + . We see this at the points East, B, F in a higher place. The value of the slope is increasing.
Now to say that the gradient is increasing, is to say that, at a critical value, the second derivative (Lesson 9) -- which is charge per unit of modify of the slope -- is positive.
Again, hither isf(x):
| f(x) | = | x 2 − 6x + five. |
| f '(10) | = | 2x − 6. |
| f ''(x) | = | two. |
f '' evaluated at the critical value 3 -- f''(3) = 2 -- is positive. This tells u.s.a. algebraically that the critical value three determines a minimum.
Sufficient weather condition
Nosotros tin can now 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 run across that at the points C, A, D.
Case 2. Letf(10) = two10 three− 9ten ii + 12x − 3.
Are there any extreme values? First, are at that place whatever 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) = half-dozenx 2 − 18x + 12 | = | half-dozen(x 2 − iiix + two) |
| = | 6(10 − i)(x − ii) | |
| = | 0 | |
implies:
x = 1 orx = 2.
(Lesson 37 of Algebra.)
Those are the critical values. Does each 1 determine a maximum or does it determine a minimum? To reply, we must evaluate the second derivative at each value.
| f '(x) | = | half-dozenx 2 − 18x + 12. |
| f ''(x) | = | 1210 − 18. |
| f ''(1) | = | 12 − xviii = −half-dozen. |
The second derivative is negative. The function therefore has a maximum at x = ane.
To detect the y-coördinate -- the farthermost value -- at that maximum we evaluatef(1):
| f(x) | = | 2x 3− nineten 2 + 12x − 3 |
| f(1) | = | 2 − 9 + 12 − 3 |
| = | two. | |
The maximum occurs at the point (ane, 2).
Next, does x = 2 make up one's mind 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 ten = ii.
To find the y-coördinate -- the extreme value -- at that minimum, nosotros evaluate f(2):
| f(x) | = | twox three − 9x 2 + 12x − three. |
| f(two) | = | 16 − 36 + 24 − 3 |
| = | ane. | |
The minimum occurs at the point (2, i).
Here in fact is the graph off(x):
Solutions to f ''(ten) = 0 indicate a betoken of inflection at those solutions, non a maximum or minimum. An case is y =x 3.y'' = 6x = 0 implies x = 0. Merely 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. Detect the coördinates of the vertex of the parabola,
y = x two − 8x + 1.
To see the answer, laissez passer your mouse over the colored surface area.
To cover the answer once again, click "Refresh" ("Reload").
Practise the problem yourself first!
y' = 2x − 8 = 0.
That implies x = 4. That's the x-coördinate of the vertex. To detect the y-coördinate, evaluate y at x = four:
y = 42 − 8· four + 1 = −15.
The vertex is at (4, −15).
Problem 2. Examine each role for maxima and minima.
a) y = x 3 − 310 2 + ii.
y' = 3x 2 − vi10 = 3ten(x − 2) = 0 implies
ten = 0 or x = 2.
y''(x) = 6ten − 6.
y''(0) = −6.
The 2d derivative is negative. That ways there is a maximum at ten = 0. That maximum value is
y(0) = 2.
Next,
y''(two) = 12 − 6 = 6.
The second derivative is positive. That means in that location is a minimum at x = 2. That minimum value is
y(2) = 2three − iii· 22 + two = 8 − 12 + 2 = −2.
b) y = −iix 3 − 3x 2 + 12 x + 10.
At x = 1 there is a maximum of y = 17.
At ten = −2 there is a minimum of y = −10.
c) y = 2x 3 + three10 2 + 12 x − 4.
Since f '(x) = 0 has no real solutions, there are no farthermost values.
d) y = 3x iv− 4x 3 − 12ten 2 + ii.
At ten = 0 there is a maximum of y = 2.
At x = −1 in that location is a minimum of y = −3.
At x = two there is a minimum of y = −xxx.
Next Lesson: Applications of maximum and minimum values
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