How do you find the critical values of a function f(x)=12x^3-27x^2-36x+8
by zeroing out the derivatives
a cubic will derive down to a quadratic; so you get 2 roots that are critical points for min and max values the quadratic will derive down to a linear whose root will be a critical point for concavity the other points to test are the ends of the interval if you gots one
All I was given in the problem was what I wrote.
yes, and prior to that you should have been given how to take derivatives in order to solve it.
the derivative of a sum is the sum of the derivatives the derivative of x^n = nx^(n-1) the derivative of a constant is 0 those types of rules will help you out in the long run then you might need to recall how to find the roots of an equation since those will be the x values that you need to plug into the original in order to obtain a y value to complete and ordered pair
The answer is -1/2,2
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