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  1. ) Consider the function f(x)=5−7x2,−5≤x≤1

The absolute maximum value is


and this occurs at xx equals


The absolute minimum value is


and this occurs at xx equals


2.) Decide whether the function f=4−x^2 satisfies the hypotheses of the MVT on the interval [a,b]=[−1,0]

Then find all values of cc in the interval [a,b]

c=

Verified Answer :

MissPhiladelphia

1.) The interval of the value of x is from -5 to 1, inclusive. Remember that what is asked is the absolute value, thus the sign does not matter even if you have to subtract x from 5. Thus, the maximum value would be obtained if the x is smaller, which is 1. The minimum value is obtained when x=-5.

Absolute maximum value: x = - 5
f(-5) = ║5 - 7(-5)^2║ = ║-170║=170


Absolute minimum value: x = 1
f(1) = ║5 - 7(1)^2║ = ║-2║= 2

2.) The Mean Value Theorem (MVT) applies to functions that are continuous and differentiable on the closed and open interval of a to b, respectively. Since the function is a quadratic function, MVT can be applied. Then, this means that there is a value of c which is between a and b. This could be determined using this formula according to MVT:

\(f'(c)= \frac{f(b)-f(a)}{b-a}\)

The differentiated form would be f'(x) = -2x. Then,

\(-2c = \frac{(4- 0^{2} )-(4- (-1)^{2}) }{0--1}=1\)

\(c=- \frac{1}{2}\)

Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.

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