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

Question

Giselle Price

Mathematics

8 December, 17:02

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

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Answers (1)

Know the Answer?

Kinley Gordon · Answer 1

a = - 5

b = - 1

m = - 1

Step-by-step explanation:

Piecewise function:

f (x) = - 5, x=0

f (x) = - x^2+3x+a, 0
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