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Section 10.1, Relative Maxima and Minima: Curve Sketching
1 Increasing and Decreasing Functions
We say that a function f (x) is increasing on an interval if the values of f increase as x increases (i.e. f โฒ(x) > 0).
A function f (x) is decreasing on an interval if the values of f decrease as x increases (i.e. f โฒ(x) < 0).
Example
The blue curve represents the graph of y = f (x), for some function f.
The function is increasing on the intervals (โ 6 , โ2) and (2, 6), and is decreasing on the intervals (โ 2 , 2), (โ 8 , โ6), and (6, 8). From the graph, we do not know what happens for x < โ8 or x > 8.
2 Relative Maxima and Minima
The point (x 1 , f (x 1 )) is a relative maximum point of the function f if there is an interval around x 1 on which f (x 1 ) โฅ f (x) for all x in that interval. We say that the relative maximum occurs at x = x 1 and the relative maximum is f (x 1 ).
Likewise, the point (x 2 , f (x 2 )) is a relative minimum point of the function f if there is an interval around x 2 on which f (x 2 ) โค f (x) for all x in that interval. We say that the relative minimum occurs at x = x 2 and the relative minimum is f (x 2 ).
On the graph above, the points where the relative maxima and minima occur have horizontal tangent lines, so f โฒ(x) = 0 at these points.
In general, if f has a relative maximum or a relative minimum at x = x 0 , either f โฒ(x 0 ) = 0 or f โฒ(x 0 ) is undefined. Such values of x are called critical values.
Example
Find the critical value(s) of the function f (x) = x^3 โ 2 x^2 + 1.
We are asked to find the values of x for which f โฒ(x) = 0 or f โฒ(x) is undefined.
f โฒ(x) = 3x^2 โ 4 x 0 = 3x^2 โ 4 x = x(3x โ 4)
x = 0,
3 Finding Relative Maxima and Minima: The First Deriva-
tive Test
To carry out the first derivative test for a function f , we need to find the critical values first. Let x 0 be a critical value of f.
Next, find an interval (a, b) in which x 0 is the only critical value. We need f to be continuous on the interval (a, b). Then, our function will fall into one of the following categories:
- On (a, b), if f โฒ(x) is negative to the left of x = x 0 and positive to the right of x = x 0 , then f (x 0 ) is a relative minimum. This happens because the function is decreasing as it approaches x = x 0 and increases afterward.
- On (a, b), if f โฒ(x) is positive to the left of x = x 0 and negative to the right of x = x 0 , then f (x 0 ) is a relative maximum. This happens because the function is increasing as it approaches x = x 0 and then decreases.
- If f โฒ(x) has the same sign to the left and right of x = x 0 within the interval (a, b), then f (x 0 ) is neither a relative maximum nor minimum of f. If, in addition f โฒ(x 0 ) = 0, then (x 0 , f (x 0 )) is called a horizontal point of inflection of f.
Examples
- Find the relative maxima and minima, and horizontal points of inflection of f (x) = 14 x^4 โ 2 3 x
2 x
(^2) + 1. Also state where the function is increasing and decreasing. Then use this information to sketch the graph of the function. First, we need to find the critical values of f , so we calculate the first derivative, set it equal to zero, and solve for x:
f โฒ(x) = x^3 โ 2 x^2 + x 0 = x^3 โ 2 x^2 + x = x(x^2 โ 2 x + 1) = x(x โ 1)^2 x = 0, 1
Testing the sign of the first derivative, we get that:
There is a relative minimum at x = 0 and a horizontal point of inflection at x = 1. We were asked for points, not just the x-values, so we calculate f (0) and f (1) to get that the relative minimum point is (0, 1) and the horizontal point of inflection is (1, 1312 ). Reusing the information from the above number line, we get that the function is decreasing on the interval (โโ, 0) and increasing on the intervals (0, 1) and (1, โ).
In words, the graph decreases until it reaches the point (0, 1). It then increases until it levels off at (1, 1312 ), then continues to increase.
- Find the critical values where the relative maxima and minima of y = โ 18 x^8 โ 27 x^7 โ 16 x^6 occur. First, we find the derivative and set it equal to zero to find the critical values:
yโฒ^ = โx^7 โ 2 x^6 โ x^5 0 = โx^5 (x^2 + 2x + 1) = โx^5 (x + 1)^2 x = โ 1 , 0
We then find the sign of the derivative around the critical values:
