Calculus 1 Math 151 Week 10 Rob Rahm. Theorem 1.1. Rolle s Theorem. Let f be a function that satisfies the following three hypotheses:

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1 Calculus 1 Math 151 Week 10 Rob Rahm 1 Mean Value Theorem Theorem 1.1. Rolle s Theorem. Let f be a function that satisfies the following three hypotheses: (1) f is continuous on [a, b]. (2) f is differentiable on (a, b). (3) f(a) = f(b). Then there is a number c in (a, b) such that f (c) = 0. Theorem 1.2. Mean Value Theorem Let f be a function that satisfies: (1) f is continuous on [a, b]. (2) f is differentiable on (a, b). Then there is a number c in (a, b) such that: f (c) = f(b) f(a). b a Example 1. We will illustrate this with an example. Let f(x) = x 3 x on [0, 2]. Then f(b) f(a) b a = f(2) f(0) 2 0 = 3. Now f (x) = 3x 2 1 and we want to solve 3x 2 1 = 3. This has two solution ±2/ 3. Since we only want solutions in [0, 2] then c = 2/ 3 is the number in (0, 2) with f (c) = f(2) f(0) 2 0 = 3. Solution 1. 1

2 2 How Derivatives Affect the Shape of a Graph 2.1 What does f say about f As we have discussed, the derivative of a function tells us information about the function itself. The first is the increasing/decreasing test: (a) If f (x) > 0 on an interval, then f is increasing on that interval. (b) if F (x) < 0 on an interval, then f is decreasing on that interval. Example 2. Find where f(x) = 3x 4 4x 3 12x 2 +5 is increasing and where it is decreasing. Solution 2. (1) Find f (x) = 12x 3 12x 2 24x = 12x(x 2 x 2) = 12x(x + 1)(x 2). (2) Determine where f (x) is negative and positive. To do this, we observe that the sign only depends on the factors x(x + 1)(x 2). Each factor changes its sing when it is zero. So, x changes from negative to positive at 0. (x + 1) changes from negative to positive at 1 and (x 2) changes from negative to positive at 2. Thus, we break the problem into 4 intervals: Interval x x-2 x+1 f (x) f(x) (, 1) dec ( 1, 0) inc (0, 2) dec (2, ) inc 2

3 So, f is decreasing on (, 1) and on (0, 2). And f is increasing on ( 1, 0) and on (2, ). By Fermat s theorem, we know that if f has a local extreme value at c, then f (c) = 0 or f (c) doesn t exist. Thus, all local mins and maxes will occur at a critical number, but not all critical numbers give a local min or max. The first derivative test gives a way to find local mins and maxes. Suppose that c is a critical number of a continuous function f. (1) If f changes from positive to negative at c, then f has a local max at c. (2) If f changes from negative to positive at c, then f has a local min at c. (3) If f does not change sign, then f has no local extreme value at c. Example 3. Consider the function f(x) = 3x 4 4x 3 12x above. Where does f have local mins and maxes? ny local extreme values will occur at critical numbers. Since f (x) is defined everywhere, we only need to look where f (x) = 12x(x 1)(x + 2) = 0 which happens at 0, 1, 2. According to the chart above, f changes sign from negative to positive at 1 and so there is a local min. And f changes from positive to negative at 0 and so there is a local max. Finally, f changes from negative to positive at 2 and so there is a local min there. 3

4 2.2 What does f say about f? We say a function is concave up on an interval if all the secant lines lie above the graph. Functions that are concave up look like a u. Consider the function f(x) = x 2. It is concave up on (, ). A function is concave down on an interval if all the secant lines lie below the graph. Functions that are concave down look like an upside down u. The function f(x) = x 2 is concave down on (, ). Functions like f(x) = x 3 are concave down on (, 0) and concave up on (0, ). Here are some picture of a concave up and concave down function (from the webpage may/excelcalculus/sec-4-5-secondderivativeconcavity.html). Figure 1: Concave Up Function Figure 2: Concave Down Function 4

5 Let s connect this geometric idea of concavity with derivatives. If a function is concave up, the derivative is increasing. For example, with the function f(x) = x 2, the derivative is negative to the left of 0 and positive to the right of 0. So the derivative is increasing. This means the second derivative is positive. This gives us the concavity test: (1) If f (x) > 0 for all x in an interval I, then the graph of f is concave upward on I. (2) If f (x) < 0 for all x in an interval I, then the graph of f is concave downward in I. Definition 1. A point P on a curve y = f(x) is called an inflection point if f is continuous there and the curve changes from concave up to concave down or from concave down from concave up. We can also use the concavity ideas to determine whether there is a local max or local min at a critical number. Observe that near a local max, a function must be concave down and near a local min, a function must be concave up. This gives the second derivative test: (a) if f (x) = 0 and if f (c) > 0, then f has a local max at c. (b) if f (x) = 0 and if f (c) < 0, then f has a local min at c. 5

6 Example 4. Consider the curve y = x 4 4x 3, find where it is concave up, concave down, points of inflection, local max, and local min. Solution 3. 6

7 Example 5. Sketch the graph of f(x) = x 2/3 (6 x) 1/3. Solution 4. 7

8 3 Indeterminate Forms and l Hospital s Rule Now the moment we have all been waiting for L Hospital s Rule: Theorem 3.1. Suppose f and g are both differentiable and g (x) 0 on an open interval I that contains a (except possibly at a). Suppose that: lim f(x) = 0 and lim g(x) = 0 x a x a or that lim f(x) = ± and lim g(x) = ±. x a x a Then: Example 6. Calculate lim x 1 ln x x 1. f(x) lim x a g(x) = lim x a f (x) lim x a g (x). Solution 5. 8

9 Example 7. Find lim x ln x x. Solution 6. Example 8. Find lim x 0 tan x x x 3. Solution 7. 9

10 Example 9. Find lim x 0 + x ln x Solution 8. Example 10. Find lim x 1 + ( 1 ) 1 ln x x 1 Solution 9. 10

11 Example 11. Find lim x (e x x). Solution