in the diagram above, imagine an officer noting the time in which an individual enters and exits a toll road…

in the diagram above, imagine an officer noting the time in which an individual enters and exits a toll road (via the time stamps on the receipts or stubs). the officer could issue a ticket based on the mean value theorem if the ratio of the distance traveled $(f(b)-f(a))$ to the travel time $(b - a)$ is greater than the speed limit. the mvt states that at some point between $a$ and $b$, the slope of the graph (the speed of the car at an instant) must be equal to the slope of the secant line through the endpoints. in other words, there must be some $c$ in the interval $(a,b)$ such that $f(c)=\frac{f(b)-f(a)}{b - a}$. click here to access the explore it in a new window. select scenario 1, which is about polynomials. change the function to $f(x)=-3 + 2x^{2}-2x^{3}$, and set the interval being considered to $-1,3$. (a) the two points through which the secant line is drawn are $(a,f(a))=(square)$ and $(b,f(b))=(square)$. (b) the slope of the secant line is $square$. this means the secant line --select--. (c) to the nearest hundredth, the value of $c$ that satisfies the conclusion of the mean value theorem in this case is $c=square$. (d) at $x = -1$, the tangent line to the function being considered has the same slope as the secant line. however, $-1$ does not satisfy the conclusion of the mean value theorem on the interval. this is because the mean value theorem guarantees the secant line over a closed interval $a,b$ is the value of the derivative at some point $c$ --select--.

in the diagram above, imagine an officer noting the time in which an individual enters and exits a toll road (via the time stamps on the receipts or stubs). the officer could issue a ticket based on the mean value theorem if the ratio of the distance traveled $(f(b)-f(a))$ to the travel time $(b - a)$ is greater than the speed limit. the mvt states that at some point between $a$ and $b$, the slope of the graph (the speed of the car at an instant) must be equal to the slope of the secant line through the endpoints. in other words, there must be some $c$ in the interval $(a,b)$ such that $f(c)=\frac{f(b)-f(a)}{b - a}$. click here to access the explore it in a new window. select scenario 1, which is about polynomials. change the function to $f(x)=-3 + 2x^{2}-2x^{3}$, and set the interval being considered to $-1,3$. (a) the two points through which the secant line is drawn are $(a,f(a))=(square)$ and $(b,f(b))=(square)$. (b) the slope of the secant line is $square$. this means the secant line --select--. (c) to the nearest hundredth, the value of $c$ that satisfies the conclusion of the mean value theorem in this case is $c=square$. (d) at $x = -1$, the tangent line to the function being considered has the same slope as the secant line. however, $-1$ does not satisfy the conclusion of the mean value theorem on the interval. this is because the mean value theorem guarantees the secant line over a closed interval $a,b$ is the value of the derivative at some point $c$ --select--.

Answer

Explanation:

Step1: Find (f(a)) and (f(b))

Given (a=-1), (f(-1)=-3 + 2(-1)^2-2(-1)^3=-3 + 2+2 = 1). Given (b = 3), (f(3)=-3+2\times3^2-2\times3^3=-3 + 18-54=-39). So ((a,f(a))=(-1,1)) and ((b,f(b))=(3,-39)).

Step2: Calculate the slope of the secant line

The slope formula is (m=\frac{f(b)-f(a)}{b - a}). Substitute (a=-1), (b = 3), (f(a)=1), (f(b)=-39) into the formula: (m=\frac{-39 - 1}{3-(-1)}=\frac{-40}{4}=-10).

Step3: Find the derivative of (f(x))

Differentiate (f(x)=-3 + 2x^2-2x^3) using the power rule ((x^n)^\prime=nx^{n - 1}). (f^\prime(x)=4x-6x^2).

Step4: Solve for (c) using the Mean - Value Theorem

Set (f^\prime(c)=\frac{f(b)-f(a)}{b - a}). So (4c-6c^2=-10). Rearrange to (6c^2-4c - 10 = 0), divide by (2) to get (3c^2-2c - 5 = 0). Factor (3c^2-2c - 5=(3c - 5)(c + 1)=0). Solutions are (c=-1) or (c=\frac{5}{3}\approx1.67). But (c=-1) is an endpoint, so (c=\frac{5}{3}\approx1.67).

Answer:

(a) ((-1,1)) and ((3,-39)) (b) (-10) (c) (1.67) (d) in the open interval ((a,b))