2,3 69, 3 xx fx xx is (A) undefined. (B) continuous but not differentiable. (C) differentiable but not continuous. (D) neither continuous nor differentiable. (E) both continuous and differentiable. (calculator not allowed) f(x) x 2if x 3 4x 7if x 3 Let f be the function given above. Math AP®︎/College Calculus AB Differentiation: definition and basic derivative rules Connecting differentiability and continuity: determining when derivatives do and do not exist Differentiability and continuity.
Show that the function has a solution between 2 and 3.
Plugging in 2 and 3 into f(x), we see that f(2) = ln(2) 1 -0.307 and f(3) = ln(3) 1 0.099. Since the first number is negative and the second number is positive and f(x) is a continuous function on the interval [2, 3], by the Intermediate Value Theorem, f(x) must have a solution between 2 and 3.
Show that the function has only one real solution.
2.3 Differentiabilityap Calculus Test
First we use the Intermediate Value Theorem to show that there is at least one solution. We can use the theorem since f(x) is a continuous function everywhere. Notice that f(1) = -1 and f(2) = 45. That means that somewhere between 1 and 2, f(x) = 0.
Well, we have shown that there is at least one solution to the equation. Now we have to show that it is the only solution. To do that, we shall show that it is not possible for f(x) to have a second solution.
We do this by taking the derivative. . Notice that . That means that f(x) is a strictly increasing function. A strictly increasing function will only have one x-intercept (solution), and thus f(x) has only one real solution.
Problems For You To Solve
1. In what situation f'(x) might NOT fail to exist？
2. Let f be a continuous function on the closed interval [-3.6]. If f (-3) =- 1 and f (6) = 3, then the Intermediate Value Theorem guarantees that
B. for at least one c between -3 and 6
C. for all x between -3 and 6
D. f(c)=1 for at least one c between -3 and 6
E. f(c)=0 for at least one c between -1 and 3
3. The function f is continuous on the closed interval [0, 2] and has values that are given in the table above. The equation must have at least two solutions in the interval [0, 2] if k=
4. Let f be a function that is differentiable on the open interval (1, 10). If f(2) = -5, f(5) = 5, and f(9) = -5, which of the following must be true?
I. f has at least 2 zeros.
II. The graph of f has at least one horizontal tangent.
III. For some c, 2 < c < 5 , f(c)=3.
B. I only
C. I and II only
D. I and III only
E. I, II and III
5. The function f is continuous for and differentiable for -2 < x < 1. If f(-2) =-5 and f(l) = 4, which of the following statements could be false?
A. There exists c, where -2 < c < 1, such that f (c) = 0.
B. There exists c, where -2 < c < 1, such that f '(c) = 0.
C. There exists c, where -2 < c < 1, such that f(c) =3.
D. There exists c, where -2 < c < 1, such that f '(c) = 3.
6. Let f(x) = x x and g(x) =sin x. Assertion : gof is differentiable at x = 0 and its derivative is continuous at that point Reason : gof is twice differentiable at x = 0.
2.3 Differentiabilityap Calculus Notes
|(I)||f(x) = x¾||for all x in [-1,1].|
|(II)||g(x) = x-1||for all x in [-1,1].|
|(III)||h(x) = x/(1-x2)||for all x in (-1,1).|
|(IV)||k(x) = 1- x 3||for all x in [-1,½].|