Write a linear function f with f(-3) =-2 and f(-2) = 3 .
f(x) = 1

Write a linear function f with the values f(3)=−2 and f(-2)=3. f(x)=1

So, this is just a different way to say two different coordinates (x, y) and (x1, y1).  Since, f(x)=y and f(x1) = y1

We have: (2,-2) and (1,1) and we want to know the equation of the line that goes through those two points.

Find the slope: M = (y-y1)/(x - x1) = (-2 - 1) / (2 - 1) = -3/1 = 3; M = 3

Find the Y intercept by plugging in one of the points for x and y and solving for B

Y = 3(X) + B

I'll choose the (1,1) to plug in: 1 = 3(1) + B; so B = -2

Put them together in slope intercept form: Y = 3(X) -2

Does this help you understand the steps to take to solve this type of problem? If not message me on here back...

answer: 48

just remember to use pemdas:

parentheses

expressions

multiplication

division

addition

subtraction

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.