relative extrema
x^3 + 1/2x^2 - 2x + 5
Ok I have
3x^2 + x - 2 = 0
Am I on the right track?
I think i know why I kept on asking for help with these pribelms. I forgot the step of factoring.
Please check...
3x^2 + x - 2 = 0
(3x-2)(x+1)=0
x=2/3, x=-1
Correct. Nice job.
If you calculate the second derivative (6x + 1) at those points, you can separate maxima from minima. The second derivative is negative at x = -1, making it a relative maximum.
To determine the relative extrema of a function, you need to find the critical points, which are the points where the derivative is either zero or undefined. In this case, you correctly found the critical points by setting the derivative of the function equal to zero:
3x^2 + x - 2 = 0
To further simplify this equation, you can try factoring it:
(3x - 2)(x + 1) = 0
Setting each factor equal to zero, you get:
3x - 2 = 0 or x + 1 = 0
Solving for x in each case gives you:
x = 2/3 or x = -1
So, the critical points are x = 2/3 and x = -1.
Next, you need to determine whether these critical points are relative maxima or minima. To do this, you can analyze the concavity of the function by finding the second derivative. The second derivative is the derivative of the first derivative:
f''(x) = (d^2)/(dx^2) (3x^2 + x - 2)
Taking the derivative gives you:
f''(x) = 6x + 1
Now, substitute each critical point into the second derivative equation to determine their concavity:
f''(2/3) = 6(2/3) + 1 = 4
f''(-1) = 6(-1) + 1 = -5
Since the second derivative is positive at x = 2/3 and negative at x = -1, the critical point x = -1 corresponds to a relative maximum, while x = 2/3 does not correspond to any type of extremum.
So, the relative extrema of the function x^3 + 1/2x^2 - 2x + 5 are:
Relative maximum at x = -1