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guendyg04
12.10.2019 •
Mathematics
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=
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Ответ:
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.
Ответ:
6×4=24
24×6=144
(no idea what to type for 20ccharavters long.)