Asked by help please
What is the minimum value of the given function?
𝑓(𝑥) = 6𝑥^2 − 12𝑥 + 𝑘, 0 ≤ 𝑥 ≤ 10, 𝑘 is a constant.
(A)𝑓(0) (B) 𝑓(10) (C) 𝑓(0) + 𝑘 (D) 𝑓(1) + 𝑘 (E) none of the above.
𝑓(𝑥) = 6𝑥^2 − 12𝑥 + 𝑘, 0 ≤ 𝑥 ≤ 10, 𝑘 is a constant.
(A)𝑓(0) (B) 𝑓(10) (C) 𝑓(0) + 𝑘 (D) 𝑓(1) + 𝑘 (E) none of the above.
Answers
Answered by
mathhelper
f(x) = 6x^2 - 12x + k
f ' (x) = 12x -12
= 0 for a max/min
12x-12=0
x = 1
f(1) = 6 - 12 + k = k - 6 , which I don't see in your choices, so ....
One of the choices is f(1) + k
which would result in
(k - 6) + k
= -6
suppose it had been f(x) = 6x^2 - 12x + 5, were k = 5
then f(1) = 6 - 12 + 5 = -1 <------ min value of the function
the choice f(1) + 5 would give us 3, which is not the minimum value
f ' (x) = 12x -12
= 0 for a max/min
12x-12=0
x = 1
f(1) = 6 - 12 + k = k - 6 , which I don't see in your choices, so ....
One of the choices is f(1) + k
which would result in
(k - 6) + k
= -6
suppose it had been f(x) = 6x^2 - 12x + 5, were k = 5
then f(1) = 6 - 12 + 5 = -1 <------ min value of the function
the choice f(1) + 5 would give us 3, which is not the minimum value
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