Question

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

Answer 1

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.