- #1
karush
Gold Member
MHB
- 3,269
- 5
So
$a(t)=v'(t)
=\dfrac{\dfrac{(50-30)mi}{h}}{\dfrac{h}{6}}
=\dfrac{20 mi}{h}\cdot\dfrac{6}{h}=\dfrac{120 mi}{h^2}$
Hopefully 🕶
Last edited by a moderator:
karush said:$10 min = \dfrac{h}{6}$
So
$a(t)=v'(t)
=\dfrac{\dfrac{(50-30)mi}{h}}{\dfrac{h}{6}}
=\dfrac{20 mi}{h}\cdot\dfrac{6}{h}=\dfrac{120 mi}{h^2}$
Hopefully 🕶
The mean value theorem is a fundamental theorem in calculus that states that if a function is continuous on a closed interval and differentiable on the open interval, then there exists at least one point in the open interval where the derivative of the function is equal to the average rate of change of the function over the closed interval.
The mean value theorem is used in a variety of real-world applications, such as in physics to calculate the average velocity of an object over a certain period of time, in economics to determine the average rate of change of a market trend, and in engineering to find the maximum and minimum values of a function.
The mean value theorem and the intermediate value theorem are both fundamental theorems in calculus, but they have different applications. The mean value theorem is used to find the average rate of change of a function over a closed interval, while the intermediate value theorem is used to show the existence of a root or solution to a function within a given interval.
No, the mean value theorem can only be applied to functions that are continuous on a closed interval and differentiable on the open interval. If a function is not continuous or differentiable, then the mean value theorem cannot be used to find the average rate of change of the function.
The mean value theorem can be interpreted geometrically as stating that for any continuous and differentiable function on a closed interval, there exists at least one point where the slope of the tangent line is equal to the slope of the secant line connecting the endpoints of the interval. This point is known as the "mean" or average point.