Question

Find the absolute extrema of the function on the closed interval

f(x)= 1 - | t -1|, [-7, 4]

minimum =

maximim =

f(x)= x^3 - (3/2)x^2, [-3, 2]

minimim =

maximim =

f(x)= 7-x, [-5, 5]

minimim =

maximim =

Answer #1

Find the absolute extrema of the function on the closed
interval. (Order your answers from smallest to largest x,
then from smallest to largest y.)
f(x) = sin(4x), [0, π]
minimum(x, y)=
(x, y)=
maximum(x, y)=
(x, y)=

Find the absolute extrema of the function on the closed
interval. h(s) = 10 / (s − 3) , [0, 1]

Find the absolute minimum and maximum of the function
on the given closed interval.
f(x)=3x4
-4x3-12x2+1 on [-2, 3]

Find the absolute maximum and minimum values on the closed
interval [-1,8] for the function below. If a maximum or minimum
value does not exist, enter NONE.
f(x) = 1 − x^2/3
Find the absolute maximum and absolute minimum values of the
function below. If an absolute maximum or minimum does not exist,
enter NONE.
f(x) = x3 - 12x on the
closed interval [-3,5]

Find absolute minimum and maximum of the function
F(x)=x^3+x^2-x+2 on closed interval (-3,0)

Consider the function f(x)=−8x−(2888/x) on the interval [11,20].
Find the absolute extrema for the function on the given interval.
Express your answer as an ordered pair (x,f(x)). (Round your
answers to 3 decimal places.)

Find the absolute extrema of the given function in the given
interval
f(x) = 4x3+3x2-18x+3 , ( 1/2,3)

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

1) find the
absolute extrema of function f(x) = 2 sin x + cos 2x on the
interval [0, 2pi]
2)
is f(x) = tanx
concave up or concave down at x = phi / 6

Find the absolute extrema of the function on the interval [1,
5]. ?(?) = 2?2 − 8? − 1

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