A function is given. What is the minimum value of the function?
π(π₯)=3π₯2 β11π₯+πΆ, 5β€π₯β€10, πΆ is a constant
2 answers
Itβs 3x^2
π(π₯)=3π₯2 β11π₯+πΆ
f ' (x) = 6x - 11
for a min of f(x), 6x - 11 = 0
x = 11/6 , which lies within the given domain
f(11/6) = 3(121/36) - 11(11/6) + C
= -121/12 + C
or
find the vertex using pre-Calculus method
y = 3(x^2 - (11/3)x + + 121/36 - 121/36) + C
= 3(x - 11/6)^2 - 121/12 + C
so the vertex is (11/6 , -121/12 + C)
Since the parabola opens upwards, there is a minimum of
-121/12 , and it occurs when x = 11/6
f ' (x) = 6x - 11
for a min of f(x), 6x - 11 = 0
x = 11/6 , which lies within the given domain
f(11/6) = 3(121/36) - 11(11/6) + C
= -121/12 + C
or
find the vertex using pre-Calculus method
y = 3(x^2 - (11/3)x + + 121/36 - 121/36) + C
= 3(x - 11/6)^2 - 121/12 + C
so the vertex is (11/6 , -121/12 + C)
Since the parabola opens upwards, there is a minimum of
-121/12 , and it occurs when x = 11/6