Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Let f(x)=(1)/(|x|). Can we use the mean value theorem to say the equation f^(')(x)=-(1)/(4) has a solution where 2 < x < 4 ? Choose 1 answer: (A) No, since the function is not differentiable on that interval. (B) No, since the average rate of change of f over the interval 2 <= x <= 4 isn't equal to -(1)/(4). (C) Yes, both conditions for using the mean value theorem have been met.

Let f(x)=1∣x∣ f(x)=\frac{1}{|x|} .\newlineCan we use the mean value theorem to say the equation f′(x)=−14 f^{\prime}(x)=-\frac{1}{4} has a solution where \( 2

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 1right chevron icon
Power rule with rational exponentsright chevron icon

Full solution

Q. Let f(x)=1∣x∣ f(x)=\frac{1}{|x|} .\newlineCan we use the mean value theorem to say the equation f′(x)=−14 f^{\prime}(x)=-\frac{1}{4} has a solution where 2<x<4 2<x<4 ?\newlineChoose 11 answer:\newline(A) No, since the function is not differentiable on that interval.\newline(B) No, since the average rate of change of f f over the interval 2≤x≤4 2 \leq x \leq 4 isn't equal to −14 -\frac{1}{4} .\newline(C) Yes, both conditions for using the mean value theorem have been met.
  1. Mean Value Theorem Conditions: The Mean Value Theorem states that if a function ff is continuous on the closed interval [a,b][a, b] and differentiable on the open interval (a,b)(a, b), then there exists at least one cc in the interval (a,b)(a, b) such that f′(c)f'(c) is equal to the average rate of change of ff over [a,b][a, b]. We need to check if f(x)=1∣x∣f(x) = \frac{1}{|x|} meets these conditions on the interval [2,4][2, 4].
  2. Continuity Check: First, we check if f(x)f(x) is continuous on the closed interval [2,4][2, 4]. Since ∣x∣|x| is never zero for xx in [2,4][2, 4], f(x)=1∣x∣f(x) = \frac{1}{|x|} is continuous on [2,4][2, 4].
  3. Differentiability Check: Next, we need to check if f(x)f(x) is differentiable on the open interval (2,4)(2, 4). The function f(x)=1∣x∣f(x) = \frac{1}{|x|} is not differentiable at x=0x = 0, but since 00 is not in the interval (2,4)(2, 4), we need to consider the differentiability elsewhere. The absolute value function ∣x∣|x| causes a problem at x=0x = 0, but since we are considering the interval (2,4)(2, 4), where xx is always positive, we can write f(x)f(x) as (2,4)(2, 4)11 for this interval. The derivative of (2,4)(2, 4)11 is (2,4)(2, 4)33, which is defined for all xx in (2,4)(2, 4).
  4. Average Rate of Change Calculation: However, we must remember that the Mean Value Theorem requires the function to be differentiable on the entire open interval a,ba, b. The function f(x)=1∣x∣f(x) = \frac{1}{|x|} is not differentiable at x=0x = 0, and although x=0x = 0 is not in the interval 2,42, 4, we must consider the behavior of the function at the endpoints to ensure there are no points of non-differentiability. Since ∣x∣|x| is not differentiable at x=0x = 0, and f(x)f(x) involves ∣x∣|x|, we must conclude that f(x)f(x) is not differentiable at x=0x = 0. However, this does not affect the differentiability on the interval 2,42, 4 since x=0x = 0 is not within this interval.
  5. Conclusion: Now, we calculate the average rate of change of f(x)f(x) over the interval [2,4][2, 4]. The average rate of change is given by f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}, where a=2a = 2 and b=4b = 4. So, we have:\newlinef(4)=1∣4∣=14f(4) = \frac{1}{|4|} = \frac{1}{4}\newlinef(2)=1∣2∣=12f(2) = \frac{1}{|2|} = \frac{1}{2}\newlineThe average rate of change is:\newline14−124−2=−142=−18\frac{\frac{1}{4} - \frac{1}{2}}{4 - 2} = \frac{-\frac{1}{4}}{2} = -\frac{1}{8}
  6. Conclusion: Now, we calculate the average rate of change of f(x)f(x) over the interval [2,4][2, 4]. The average rate of change is given by f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}, where a=2a = 2 and b=4b = 4. So, we have:\newlinef(4)=1∣4∣=14f(4) = \frac{1}{|4|} = \frac{1}{4}\newlinef(2)=1∣2∣=12f(2) = \frac{1}{|2|} = \frac{1}{2}\newlineThe average rate of change is:\newline14−124−2=−142=−18\frac{\frac{1}{4} - \frac{1}{2}}{4 - 2} = \frac{-\frac{1}{4}}{2} = -\frac{1}{8}The average rate of change of ff over the interval [2,4][2, 4] is [2,4][2, 4]00, which is not equal to [2,4][2, 4]11. Therefore, the Mean Value Theorem cannot guarantee that there is a number [2,4][2, 4]22 in the interval [2,4][2, 4]33 such that [2,4][2, 4]44 because the average rate of change is not equal to [2,4][2, 4]11.

More problems from Power rule with rational exponents

Question
Simplify.\newline811281^{\frac{1}{2}}\newline_____________
Get tutor helpright-arrow

Posted 9 months ago

Question
Simplify.\newline(−32)1/5(-32)^{1/5}\newline______
Get tutor helpright-arrow

Posted 9 months ago

Question
Simplify. Assume all variables are positive.\newlinew43w73\frac{w^{\frac{4}{3}}}{w^{\frac{7}{3}}}\newlineWrite your answer in the form AA or AB\frac{A}{B}, where AA and BB are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.\newline______
Get tutor helpright-arrow

Posted 5 months ago

Question
Simplify. Assume all variables are positive.\newline(2r13)8(2r^{\frac{1}{3}})^8\newline\newlineWrite your answer in the form AA or AB\frac{A}{B}, where AA and BB are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.\newline______
Get tutor helpright-arrow

Posted 5 months ago

Question
Simplify. Assume all variables are positive.\newlinev74⋅v94v^{\frac{7}{4}} \cdot v^{\frac{9}{4}}\newline\newlineWrite your answer in the form AA or AB\frac{A}{B}, where AA and BB are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.\newline______\newline
Get tutor helpright-arrow

Posted 5 months ago

Question
Simplify.\newline114⋅1141^{\frac{1}{4}} \cdot 1^{\frac{1}{4}}\newline______
Get tutor helpright-arrow

Posted 5 months ago

Question
Simplify. \newline32(−1/5)32^{(-1/5)}
Get tutor helpright-arrow

Posted 7 months ago

Question
Simplify. Assume all variables are positive.\newlinew32w52\frac{w^{\frac{3}{2}}}{w^{\frac{5}{2}}}\newline\newlineWrite your answer in the form AA or AB\frac{A}{B}, where AA and BB are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.\newline______\newline
Get tutor helpright-arrow

Posted 9 months ago

Question
Simplify. Assume all variables are positive.\newline(27x)23(27x)^{\frac{2}{3}}\newline\newlineWrite your answer in the form AA or AB\frac{A}{B}, where AA and BB are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.\newline______
Get tutor helpright-arrow

Posted 8 months ago

Question
Simplify. Assume all variables are positive.\newlinex14x114\frac{x^{\frac{1}{4}}}{x^{\frac{11}{4}}}\newline\newlineWrite your answer in the form AA or AB\frac{A}{B}, where AA and BB are constants or variable expressions that have no variables in common. All exponents in your answer should be positive.\newline______\newline
Get tutor helpright-arrow

Posted 7 months ago

View all (45)

Related topics

Algebra - Order of OperationsAlgebra - Distributive Property`X` and `Y` AxesGeometry - Scalene TriangleCommon MultipleGeometry - Quadrant