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Mean Value Theorem with Maximum Value

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Wize University Calculus 1 Textbook > Applications of Differentiation

The Mean Value Theorem

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If f(x)f(x) is continuous and differentiable on [6,15][6,15] with f(6)=−2f(6)=-2 and f′(x)≤10f^{\prime}(x)\leq 10, what is the largest possible value for f(15)f(15)?


More The Mean Value Theorem Questions:
Practice: MVT
For the function g(x)=(x+1)3g(x)=(x+1)^{3} on [−1,1][-1,1], find the number cc guaranteed by the Mean Value Theorem.
Mean Value Theorem with Maximum Value
If f(x)f(x) is continuous and differentiable on [6,15][6,15] with f(6)=−2f(6)=-2 and f′(x)≤10f^{\prime}(x)\leq 10, what is the largest possible value for f(15)f(15)?
The Mean Value Theorem
For g(x)=−x2g(x)=-x^{2} on [−1,2][-1,2], find the number cc guaranteed by the Mean Value Theorem.
Suppose that ff is a differentiable function such that f′(x)≥3f'(x)\ge3 for all xx. If f(2)=3,f(2)=3, what is the minimum possible value of f(4)f(4)?
Mean Value Theorem: Speed
In 2 hours, a car driver covered 150 km on a road with speed limit 65 km per hour. Did the car go over the speed limit? Answer using the mean-value theorem.
Mean Value Theorem with Maximum Value
If f(x)f(x) is continuous and differentiable on [6,15][6,15] with f(6)=−2f(6)=-2 and f′(x)≤10f^{\prime}(x)\leq 10, what is the largest possible value for f(15)f(15)?
Practice: MVT
Q:\textbf{Q:} For the function g(x)=(x+1)3g(x)=(x+1)^{3} on [−1,1][-1,1], find the number cc guaranteed by the Mean Value Theorem.
The Mean Value Theorem
Let f(x)f(x) be a function so that
f(x),f′(x),f′′(x)f(x), f'(x), f''(x) exist and are continuous for all xx, and
∣f(x)−sin⁡(x)∣≤1/4|f(x) - \sin(x)| \leq 1/4 for all xx.
The Mean Value Theorem
Let f(x)f(x) be a continuous function defined for all real numbers xx. Suppose f(x)f(x) is increasing on the intervals (−∞,−2)(-\infty, -2) and (4,∞)(4, \infty), decreasing on (−2,4)(-2, 4), f(−2)=5f(-2) = 5 and f(4)=1f(4) = 1. How many zeros does f(x)f(x) have?
The Mean Value Theorem
Why doesn't the function f(x)=∣x∣f(x) = \left|x\right|on the interval [−1,1][-1,1]violate the Mean Value Theorem?
In other words, how is it that f(1)−f(−1)1−(−1)=0\dfrac{f(1)-f(-1)}{1-(-1)}=0, but there does not exist a value of ccsuch that f′(c)=0f'(c) = 0?
Given f(x)=x2−3x+2f(x) = x^2 - 3x + 2on the interval x∈[−1,3]x \in [-1,3], what value of cc does the mean value theorem guarantee exists?
Suppose that ff is a differentiable function such that f′(x)≥3f'(x)\ge3 for all xx. If f(2)=3,f(2)=3, what is the minimum possible value of f(4)f(4)?
Given f(x)=x2−3x+2f(x) = x^2 - 3x + 2on the interval x∈[−1,3]x \in [-1,3], what value of cc does the mean value theorem guarantee exists?
tan question
Given the function f(x)=tan⁡xf(x)=\tan x. By the Mean Value Theorem, there exists a cc in the interval (1,2)(1,2) such that f′(c)f'(c) is equal to f(2)−f(1)2−1\dfrac{f(2)-f(1)}{2-1} . This statement is:
Practice: Mean Value Theorem
Practice: Mean Value Theorem
Given f(x)=x2−3x+2f(x) = x^2 - 3x + 2on the interval x∈[−1,3]x \in [-1,3], what value of cc does the mean value theorem guarantee exists?
Practice: Mean Value Theorem
Practice: Mean Value Theorem
If f(x)f(x) is continuous and differentiable on [6,15][6,15] with f(6)=−2f(6)=-2and f′(x)≤10f^{\prime}(x)\leq 10, what is the largest possible value for f(15)f(15)?
Given f(x)=x2−3x+2f(x) = x^2 - 3x + 2on the interval x∈[−1,3]x \in [-1,3], what value of cc does the mean value theorem guarantee exists?
Mean Value Theorem: Speed
In 2 hours, a car driver covered 150 km on a road with speed limit 65 km per hour. Did the car go over the speed limit? Answer using the mean-value theorem.
Practice: MVT
For the function g(x)=(x+1)3g(x)=(x+1)^{3} on [−1,1][-1,1], find the number cc guaranteed by the Mean Value Theorem.
Mean Value Theorem with Maximum Value
If f(x)f(x) is continuous and differentiable on [6,15][6,15] with f(6)=−2f(6)=-2 and f′(x)≤10f^{\prime}(x)\leq 10, what is the largest possible value for f(15)f(15)?
The Mean Value Theorem
For g(x)=−x2g(x)=-x^{2} on [−1,2][-1,2], find the number cc guaranteed by the Mean Value Theorem.
Example: Mean Value Theorem
Is there a constant cc in f(x)=∣x−1∣f(x)=| x-1| where f(3)−f(0)=f′(c)(3−0)f(3)-f(0)=f'(c)(3-0)? Why does this not contradict the Mean Value Theorem?
Exercise
Suppose ff is a differentiable function such that f′(x)≤5f'(x)\leq 5 for all x∈[1,5]x\in[1,5]and f(1)=3f(1)=3. What is an upper bound on f(5)f(5)?
The Mean Value Theorem
If f(x)f(x) is continuous and differentiable on [6,15][6,15] with f(6)=−2f(6)=-2and f′(x)≤10f^{\prime}(x)\leq 10, what is the largest possible value for f(15)f(15)?
Suppose that ff is a differentiable function such that f′(x)≥3f'(x)\geq 3 for all xx. If f(2)=3f(2)=3, what is the minimum possible value of f(4)f(4) ?
Given f(x)=x2−3x+2f(x) = x^2 - 3x + 2on the interval x∈[−1,3]x \in [-1,3], what value of cc does the mean value theorem guarantee exists?