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

Answer 2

The value of a is -5,m is -1 and b=-1

We have given that,

$f(x)= -x^{2} +3x+a, 0 < x < 2$

What is the meaning of Piece wise function?

A function which is defined by multiple sub-functions, each sub-function applying to a certain interval of the main function's domain

$f(x)= -5, x=0$

$f(x)= -x^{2} +3x+a, 0 < x < 2$

$f(x)= mx + b ... 2\leq x\leq 3$

Now by using the mean value theorem 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

Find the derivative of f(x)

$\frac{d}{dx}(-x^{2} +3x-5)=-2x+3$

and derivative of $mx + b$ is m, at x = 2

Therefore we get,

-2(2) + 3 = m

-1 = m

Replacing x by 2 in the functions

$-x^{2} +3x-5 = -x + b$

$-2^{2} +3*2-5 = -2 + b$

$-3 + 2 = b$

$-1 = b$

Therefore, the value of a=-5,m=-1 and b=-1

to learn more about the piece wise function visit:

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