how do you find the maximum value of a function
10
MAXIMUM AND MINIMUM
VALUES
The turning points of a graph
Westward E SAY THAT A Office f(ten) 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 considering other values of the office may in fact exist greater.
We say that a role f(10) has a relative minimum value at x = b,
if f(b) is less than whatsoever value immediately preceding or follwing.
Once again, other values of the function may in fact be less. With that agreement, then, we will drib the term relative.
The value of the part, the value of y, at either a maximum or a minimum is called an extreme value.
Now, what characterizes the graph at an extreme value?
The tangent to the curve is horizontal. We see 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 signalC -- 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 '(x) changes sign from + to − .
At a minimum, f '(x) changes sign from − to + . We can see that at the points Eastward and F.
We can as well notice that at a maximum, at A, the graph is concave downward. (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 chosen a critical value. In the figure --
-- the critical values are ten =a and x =b.
The critical values determine turning points, at which the tangent is parallel to the x-axis. The critical values -- if any -- volition be the solutions tof '(x) = 0.
Example one. Letf(x) = x 2 − vi10 + 5.
Are there any disquisitional values -- whatsoever turning points? If then, 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) = 2x − vi = 0 impliesx = iii. (Lesson ix of Algebra.)
x = 3 is the only critical value. It is the ten-coördinate of the turning signal. To determine the y-coördinate, evaluate f at that critical value -- evaluate f(3):
| f(x) | = | ten 2 − 6x + 5 |
| f(iii) | = | 32 − six· 3 + five |
| = | −4. | |
The extreme value is −four. To run across whether it is a maximum or a minimum, in this case we tin only look at the graph.
f(x) is a parabola, and nosotros can run into that the turning point is a minimum.
Past finding the value of x where the derivative is 0, then, we have discovered that the vertex of the parabola is at (iii, −four).
But we will not always exist 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 East, B, F above. The value of the gradient is increasing.
Now to say that the slope is increasing, is to say that, at a critical value, the second derivative (Lesson nine) -- which is rate of change of the slope -- is positive.
Again, hither isf(x):
| f(x) | = | x ii − 6x + v. |
| f '(x) | = | 2x − 6. |
| f ''(x) | = | 2. |
f '' evaluated at the disquisitional value 3 -- f''(three) = 2 -- is positive. This tells us algebraically that the disquisitional value iii determines a minimum.
Sufficient conditions
We can at present country these sufficient conditions for extreme values of a function at a critical value a:
The office has a minimum value at x =a if f '(a) = 0
and f ''(a) = a positive number.
The role has a maximum value at x =a if f '(a) = 0
and f ''(a) = a negative number.
In the example of the maximum, the slope of the tangent is decreasing -- information technology is going from positive to negative. We can see that at the points C, A, D.
Example 2. Letf(x) = iix iii− ninex ii + 12x − 3.
Are there any farthermost values? First, are there any critical values -- solutions to f '(10) = 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 − 18ten + 12 | = | six(ten 2 − 3x + 2) |
| = | 6(x − 1)(x − 2) | |
| = | 0 | |
implies:
x = 1 orx = 2.
(Lesson 37 of Algebra.)
Those are the critical values. Does each one decide a maximum or does it decide a minimum? To answer, we must evaluate the 2nd derivative at each value.
| f '(ten) | = | 6x 2 − 18x + 12. |
| f ''(x) | = | 12x − 18. |
| f ''(one) | = | 12 − xviii = −6. |
The second derivative is negative. The function therefore has a maximum at x = 1.
To notice the y-coördinate -- the extreme value -- at that maximum we evaluatef(one):
| f(10) | = | 210 3− ninex 2 + 12x − 3 |
| f(1) | = | 2 − nine + 12 − 3 |
| = | 2. | |
The maximum occurs at the point (ane, 2).
Next, does ten = 2 determine a maximum or a minimum?
| f ''(10) | = | 12x − 18. |
| f ''(2) | = | 24 − 18 = vi. |
The second derivative is positive. The function therefore has a minimum at x = 2.
To notice the y-coördinate -- the farthermost value -- at that minimum, we evaluate f(2):
| f(x) | = | ii10 3 − 9x ii + 12x − 3. |
| f(2) | = | xvi − 36 + 24 − 3 |
| = | ane. | |
The minimum occurs at the point (two, 1).
Here in fact is the graph off(ten):
Solutions to f ''(x) = 0 signal a point of inflection at those solutions, not a maximum or minimum. An case is y =10 3.y'' = half-dozenx = 0 implies ten = 0. But x = 0 is a indicate of inflection in the graph of y =x iii, 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 ii − eightx + 1.
To encounter the answer, pass your mouse over the colored expanse.
To cover the answer once again, click "Refresh" ("Reload").
Do the trouble yourself first!
y' = 2x − 8 = 0.
That implies x = 4. That'southward the x-coördinate of the vertex. To observe the y-coördinate, evaluate y at x = four:
y = 4ii − 8· 4 + one = −fifteen.
The vertex is at (four, −15).
Problem 2. Examine each office for maxima and minima.
a) y = x iii − 3x 2 + 2.
y' = 3x 2 − vix = threex(x − 2) = 0 implies
x = 0 or x = two.
y''(x) = half-dozenx − 6.
y''(0) = −vi.
The second derivative is negative. That means there is a maximum at x = 0. That maximum value is
y(0) = 2.
Side by side,
y''(2) = 12 − six = 6.
The second derivative is positive. That ways in that location is a minimum at x = 2. That minimum value is
y(2) = 2three − 3· 22 + ii = 8 − 12 + 2 = −2.
b) y = −2x 3 − three10 2 + 12 ten + 10.
At x = 1 there is a maximum of y = 17.
At 10 = −2 there is a minimum of y = −10.
c) y = twox 3 + threex 2 + 12 10 − 4.
Since f '(10) = 0 has no real solutions, there are no extreme values.
d) y = threeten 4− 4x 3 − 12x ii + 2.
At x = 0 in that location is a maximum of y = 2.
At x = −ane there is a minimum of y = −iii.
At 10 = 2 there is a minimum of y = −thirty.
Next Lesson: Applications of maximum and minimum values
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