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How on the earth do you do this ????
2 answers:
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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:
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.
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:
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.
If this is supposed to be
Then this can simply be reduced to
by the double angle formula for cosine
Then this can simply be reduced to
by the double angle formula for cosine