Homework SourseFind the critical points if any of the following function on
Find the critical points, if any, of the following function on the given interval. Determine the absolute extreme values of f on the given interval. f(x)=3x^3-3x^2-3x+7 on the interval [-2,-1]
I know that -1/3 is not the critical point.
I know that -1/3 is not the critical point.
Solution
differentiate f(x) and equate it to zero to find out the critical points
f\'(x) = 9x^2 - 6x - 3 = 0
==> 3x*x - 2x - 1 = 0
3x(x-1) + (x-1) = 0
x= -1/3 ,1
both are critical points
[In calculus, a critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0.[1][2] The value of the function at a critical point is a critical value of the function. ]
1 is not i the given interval [-2,-1]
f\"(x) = 18x - 6
f\"(-1/3) = -12 (negative)
so f(x) has maximum value at x = -1/3
absolute maximum = f(-1/3) = 68/9 = 7.56
minimum occurs at one of the points -2 or -1
f(-2) = -23
f(-1) =4
so absolute minimum = -23
If you dont understand anything please let me know
