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Mean Value Theorem: Speed

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Wize University Calculus 1 Textbook > Applications of Differentiation

The Mean Value Theorem

3 Activities

Wize University Calculus 1 Textbook > Applications of Differentiation for Science

Position, Velocity, and Acceleration

3 Activities

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.




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)?
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 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?
More Position, Velocity, and Acceleration Questions:
Applications of Integration: Position, Velocity and Acceleration
If the accelertion of a particle is given by a(x)=2x+xa\left(x\right)=2^x+x, and v(0)=s(0)=0v\left(0\right)=s\left(0\right)=0, then the displacement of the particle s(x)s\left(x\right) is
Practice: Velocity and Acceleration
A point PP moves along the x-axis. Its position after tt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60t meters.
Practice: Velocity and Acceleration
Q.\textbf{Q.} A point PP moves along the x-axis. Its position after tt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60t meters.
Practice: Velocity and Acceleration
Q.\textbf{Q.} A point PP moves along the x-axis. Its position after tt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60t meters.
Applications of derivatives: Velocity and Direction
At time t≥0t\ge0, the velocity (in m/s) of a particle moving along a horizontal line is v(t)=t2−2t+2v(t)=t^2-2t+2
Applications of derivatives: Velocity and Direction
At time t≥0t\ge0, the velocity (in m/s) of a particle moving along a horizontal line is v(t)=t2−2t+2v(t)=t^2-2t+2
Written Answer 3
The velocity of a protein inside a cell starting at the top and falling under the force of gravity is given by the differential equation
dvdt=g−kv\displaystyle \frac{dv}{dt} = g - kv
where g≈107μm/s2g \approx 10^7 \mu m / s^2 is the gravitational acceleration. kvkv is the acceleration due to drag and k=1013s−1k = 10^{13} s^{-1}. (Note that this model ignores diffusion of the protein which makes it a really bad model but we're exploring what gravity alone would do to a protein.
Applications of Integration: Position, Velocity and Acceleration
If the acceleartion of a particle is given by a(x)=2x+xa\left(x\right)=2^x+x, and v(0)=s(0)=0v\left(0\right)=s\left(0\right)=0, then the displacement of the particle s(x)s\left(x\right) is
Applications of Integration: Position, Velocity and Acceleration
If the displacement of a particle is given by s(t)=∫0tx2sin⁡xdxs\left(t\right)=\int_0^t\frac{x^2}{\sin x}dx, then the acceleration at at t=π2t=\frac{\pi}{2} is
Applications of Integration: Position, Velocity and Acceleration
The velocity of a particle is given by v(t)=∫05f(x)dxv\left(t\right)=\int_0^5f\left(x\right)dx, the acceleration of this particle is
Related Rates
An object is moving in such a way that its position is given by x(t)=2t3−tx(t)=2t^3-t Find the time ttat which the object’s speed is equal to the average speed over the interval [0, 1].
Kinematics: Free fall
A ball at rest is dropped off a 150-m tall building. Ignoring air resistance, what will its speed be the instant it hits the ground?
Enter your answer as a positive value (speed) in m/s, do not include units.
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: Velocity and Acceleration
Q.\textbf{Q.} A point PP moves along the x-axis. Its position after tt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60t meters.
Applications of derivatives: Velocity and Direction
At time t≥0t\ge0, the velocity (in m/s) of a particle moving along a horizontal line is v(t)=t2−2t+2v(t)=t^2-2t+2
An object is moving according to x(t)=t4−8t2+2x(t)=t^4-8t^2+2. Find the time at which its acceleration is zero.
x(t)=6t3+3t2−3,x(t)=6t^3+3t^2-3, Find velocity and acceleration at t=2?
Applications of Integration
A particle moves in a straight line so that its velocity at time tt is v(t)=t+1v(t)=\sqrt{t+1}. If its position at t=8t=8 is s(8)=20s(8)=20, find s(9)s(9).