Question

Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 3x^2 − 18x + 8, [0, 8] absolute minimum value.

Answer #1

Find the absolute maximum and absolute minimum values of f on
the given interval.
f(x) = xe-x^2/128, [-3,16]

find the absolute maximum and absolute minimum values of f on the
given interval
f(x) = x^4-2x^2+1 [-2,3]

Find the absolute maximum and absolute minimum values of
f on the given interval.
f(x) = 4x3 −
12x2 − 36x +
6,
[−2, 4]

Find the absolute maximum and absolute minimum values of f on
the given interval: x^4-8x^2+8 [-3, 4]
Absolute minimum:
Absolute maximum:

find the absolute maximum and absolute minimum values of f on
the given closed interval
f(x)=5-x^2
[-3,1]

Find the absolute maximum and minimum values of each function
over the indicated interval, and indicate the x-values at which
they occur.
f(x) = 3x^3 - 3x^2 - 3x + 7 ; [-1 , 0]
The absolute maximum value is _ at x = _
(Use a comma to separate answers as needed. Type an integer or
a fraction.)

Find the absolute maximum and absolute minimum values of
f on the given interval.
f(t) =
t
16 −
t2
, [−1, 4]

Find the absolute maximum and absolute minimum values (if they
exist) of f(x) = ln x/x on the interval (0, ∞).

Find the absolute maximum and minimum values of f(x)=
−x^3−3x^2+4x+3, if any, over the interval
(−∞,+∞)(−∞,+∞).
I know it doesn't have absolute maxima and minima but where do
they occur? In other words x= ? for the maxima and minima?

(a) Find the maximum and minimum values of f(x) = 3x 3 − x on
the closed interval [0, 1] by the following steps:
i. Observe that f(x) is a polynomial, so it is continuous on the
interval [0, 1].
ii. Compute the derivative f 0 (x), and show that it is equal to
0 at x = 1 3 and x = − 1 3 .
iii. Conclude that x = 1 3 is the only critical number in...

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