How To Find The Minimum Value Of A Function
10
MAXIMUM AND MINIMUM
VALUES
The turning points of a graph
Westward E SAY THAT A Role f(x) has a relative maximum value at x = a,
if f(a) is greater than whatsoever value immediately preceding or follwing.
We call it a "relative" maximum because other values of the office may in fact exist greater.
We say that a part 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 role may in fact be less. With that understanding, and then, we will drop the term relative.
The value of the function, the value of y, at either a maximum or a minimum is called an extreme value.
At present, what characterizes the graph at an farthermost value?
The tangent to the curve is horizontal. Nosotros come across 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 betokenC -- the gradient of the tangent is positive:f '(ten) > 0. While at points immediately to the correct -- 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 tin see that at the points E and F.
We tin can also observe that at a maximum, at A, the graph is concave downwardly. (Topic 14 of Precalculus.) While at a minimum, at B, it is concave upward.
A value of 10 at which the function has either a maximum or a minimum is chosen a disquisitional value. In the effigy --
-- the critical values are 10 =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 -- will exist the solutions tof '(x) = 0.
Instance 1. Permitf(10) = ten two − 6x + 5.
Are there whatsoever critical values -- whatever turning points? If and then, do they decide a maximum or a minimum? And what are the coördinates on the graph of that maximum or minimum?
Solution.f '(x) = 2x − six = 0 impliesx = three. (Lesson nine of Algebra.)
x = 3 is the only critical value. It is the x-coördinate of the turning point. To determine the y-coördinate, evaluate f at that critical value -- evaluate f(3):
| f(x) | = | x 2 − 6x + 5 |
| f(3) | = | 32 − vi· 3 + 5 |
| = | −four. | |
The extreme value is −4. To see whether it is a maximum or a minimum, in this instance nosotros tin can simply look at the graph.
f(x) is a parabola, and nosotros can see that the turning indicate is a minimum.
By finding the value of x where the derivative is 0, so, 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 condition for a minimum is that f '(x) changes sign from − to + . We see this at the points Eastward, B, F above. The value of the gradient is increasing.
Now to say that the slope is increasing, is to say that, at a disquisitional value, the second derivative (Lesson 9) -- which is charge per unit of change of the slope -- is positive.
Again, here isf(ten):
| f(x) | = | ten 2 − half dozenx + 5. |
| f '(x) | = | iix − six. |
| f ''(x) | = | two. |
f '' evaluated at the critical value 3 -- f''(three) = 2 -- is positive. This tells us algebraically that the critical value 3 determines a minimum.
Sufficient conditions
Nosotros can now state these sufficient conditions for extreme values of a part 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 role 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 -- information technology is going from positive to negative. We can see that at the points C, A, D.
Example two. Letf(10) = twox 3− ixx ii + 12x − 3.
Are there any extreme values? First, are there whatever critical values -- solutions to f '(x) = 0 -- and exercise they make up one's mind 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) = 6ten 2 − 18ten + 12 | = | half-dozen(10 2 − 310 + 2) |
| = | 6(10 − one)(x − 2) | |
| = | 0 | |
implies:
10 = 1 or10 = 2.
(Lesson 37 of Algebra.)
Those are the critical values. Does each i determine a maximum or does it decide a minimum? To respond, we must evaluate the second derivative at each value.
| f '(x) | = | 6x 2 − eighteen10 + 12. |
| f ''(x) | = | 12x − 18. |
| f ''(1) | = | 12 − eighteen = −six. |
The second derivative is negative. The function therefore has a maximum at x = 1.
To detect the y-coördinate -- the extreme value -- at that maximum we evaluatef(ane):
| f(x) | = | iix three− 9x 2 + 12x − 3 |
| f(1) | = | 2 − 9 + 12 − iii |
| = | 2. | |
The maximum occurs at the point (one, 2).
Next, does ten = two decide a maximum or a minimum?
| f ''(x) | = | 12x − 18. |
| f ''(2) | = | 24 − 18 = half-dozen. |
The second derivative is positive. The function therefore has a minimum at 10 = 2.
To discover the y-coördinate -- the extreme value -- at that minimum, nosotros evaluate f(2):
| f(10) | = | twox three − 9ten 2 + 12ten − 3. |
| f(ii) | = | xvi − 36 + 24 − 3 |
| = | 1. | |
The minimum occurs at the point (ii, 1).
Here in fact is the graph off(10):
Solutions to f ''(10) = 0 point a point of inflection at those solutions, not a maximum or minimum. An example is y =x 3.y'' = 6x = 0 implies x = 0. Only ten = 0 is a betoken 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.
Trouble 1. Notice the coördinates of the vertex of the parabola,
y = x 2 − eightx + 1.
To see the answer, pass your mouse over the colored area.
To comprehend the answer again, click "Refresh" ("Reload").
Do the problem yourself commencement!
y' = twox − 8 = 0.
That implies x = 4. That'south the x-coördinate of the vertex. To find the y-coördinate, evaluate y at x = 4:
y = 4two − eight· iv + 1 = −15.
The vertex is at (4, −xv).
Problem two. Examine each part for maxima and minima.
a) y = x three − iiix 2 + 2.
y' = 3ten two − 6ten = 310(ten − 2) = 0 implies
10 = 0 or x = 2.
y''(x) = 6ten − half dozen.
y''(0) = −six.
The second derivative is negative. That means in that location is a maximum at x = 0. That maximum value is
y(0) = 2.
Side by side,
y''(two) = 12 − 6 = vi.
The second derivative is positive. That means there is a minimum at x = two. That minimum value is
y(ii) = 2three − 3· ii2 + 2 = 8 − 12 + 2 = −2.
b) y = −2ten 3 − 3ten 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 = 2ten 3 + 3ten 2 + 12 x − 4.
Since f '(x) = 0 has no real solutions, in that location are no extreme values.
d) y = 3ten iv− 4ten 3 − 12ten two + ii.
At x = 0 there is a maximum of y = 2.
At 10 = −1 there is a minimum of y = −3.
At ten = 2 there is a minimum of y = −thirty.
Next Lesson: Applications of maximum and minimum values
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