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 = - 5f(-5) = ║5 - 7(-5)^2║ = ║-170║=
170Absolute minimum value:
x = 1f(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:

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


Thus, x = -1, x = -1/2, and x=0 all lie in the function 4-x^2.
Hello,
Please write correctly.
g(x)= (2x+3)/(x+1) to be written like a/(x+p) + q
It is very simple!
(2x+3)/(x+1)= [(2x+2)+1]/(x+1)= (2(x+1)+1)/(x+1)= 2+1/(x+1)
11) 7 + -10+(4(5+-3)) = 7+-10+8=5
12) (12 +-10)+3x+-1(2x+4x), 12+-10x+3x+-2x+-4x = 12+-13x
Answer:
q = 26
Step-by-step explanation:
Given the point (3, q ) lies on the line then the coordinates satisfy the equation.
Substitute x = 3, y = q into the equation
q = 8(3) + 2 = 24 + 2 = 26
The correct answer is B. 0