Question

For what values of​ a, m, and b does the function​ f(x) satisfy the hypotheses of the mean value theorem on the interval [0 comma 3 ][0,3]​? ​f(x)equals=left brace Start 3 By 3 Matrix 1st Row 1st Column negative 5 2nd Column 3rd Column x equals 0 2nd Row 1st Column negative x squared plus 3 x plus a 2nd Column 3rd Column 0 less than x less than 2 3rd Row 1st Column mx plus b 2nd Column 3rd Column 2 less than or equals x less than or equals 3 EndMatrix −5 x=0 −x2+3x+a 0

Answer 1

Answer:

a = -5

b = -1

m = -1

Step-by-step explanation:

Piecewise function:

f(x)= -5, x=0

f(x)= −x^2+3x+a, 0<x<2

f(x)= mx + b 2≤x≤3

The mean value theorem is applicable if f(x) is continuous on the entire closed interval [0;3] and differentiable on the entire closed interval [0;3].

To satisfied that at x = 0 :

−x^2+3x+a = -5

Then, a = -5

 

Derivative of −x^2+3x-5 is -2x + 3 and derivative of mx + b is m, at x = 2:

-2(2) + 3 = m

-1 = m

Replacing the functions at x = 2 :

−x^2+3x-5 = -x + b

−(2)^2+3*2-5 = -2 + b

-3 + 2 = b

-1 = b