2.3 Differentiabilityap Calculus
(1) Find the derivatives of the following functions using first principle.
All students take calculus All sin tan cos rule. Trigonometric ratios of some negative angles. Trigonometric ratios of 90 degree minus theta. Trigonometric ratios of 90 degree plus theta. Trigonometric ratios of 180 degree plus theta. Trigonometric ratios of 180 degree minus theta. Trigonometric ratios of 180 degree plus theta. Prepare your calculus students for the topics they need to know to succeed in Calculus 2 and AP Calculus BC. Students tend to forget the PreCalculus topics they didn't use in Calculus 1 or AB. This review can be used for summer review or between semesters for Calculus BC. 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. Example 2: Show that the function has only one real solution. Live: Biden on the verge of reaching 270 electoral votes. Anne Hathaway apologizes to disability community. Did the 49ers swing a local California election?
(i) f(x) = 6Solution
(ii) f(x) = -4x + 7Solution
(iii) f(x) = -x^{2} + 2 Solution
(2) Find the derivatives from the left and from the right at x = 1 (if they exist) of the following functions. Are the functions differentiable at x = 1?
(i) f(x) = x - 1 Solution
(ii) f(x) = √(1 - x^{2}) Solution
(3) Determine whether the following function is differentiable at the indicated values.
(i) f(x) = x x at x = 0 Solution
(ii) f(x) = x^{2} - 1 at x = 1 Solution
(iii) f(x) = x + x - 1 at x = 0, 1 Solution
(iv) f(x) = sin x at x = 0 Solution
(4) Show that the following functions are not differentiable at the indicated value of x.
(i)
Solution
(5) The graph of f is shown below. State with reasons that x values (the numbers), at which f is not differentiable.
Solution
(6) If f(x) = x + 100 + x^{2}, test whether f'(-100) exists.
(7) Examine the differentiability of functions in R by drawing the diagrams.
(i) sin x Solution
(ii) cos x Solution
Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.
If you have any feedback about our math content, please mail us :
We always appreciate your feedback.
You can also visit the following web pages on different stuff in math.
ALGEBRA Negative exponents rules COMPETITIVE EXAMS APTITUDE TESTS ONLINE ACT MATH ONLINE TEST TRANSFORMATIONS OF FUNCTIONS ORDER OF OPERATIONS WORKSHEETS | TRIGONOMETRY Trigonometric identities MENSURATION GEOMETRY COORDINATE GEOMETRY CALCULATORS MATH FOR KIDS LIFE MATHEMATICS SYMMETRY CONVERSIONS |
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Trigonometry word problems
Word problems on mixed fractrions
OTHER TOPICS
Ratio and proportion shortcuts
Converting repeating decimals in to fractions
SBI!
Example 1:
Show that the function has a solution between 2 and 3.
Solution:
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.
Solution:
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
Multiple Choices:
1. In what situation f'(x) might NOT fail to exist？
A. cuspe
B. conner
C. jump
D. continuous
E. IDK
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
A. f(0)=0
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=
A. 0
B. 1
C. 2
D. 3
E. 4
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.
A. None
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 Test
2.3 Differentiabilityap Calculus Notes
A. (a+b)/2
(I) | f(x) = x^{¾} | for all x in [-1,1]. |
(II) | g(x) = x-1 | for all x in [-1,1]. |
(III) | h(x) = x/(1-x^{2}) | for all x in (-1,1). |
(IV) | k(x) = 1- x ^{3} | for all x in [-1,½]. |