Again, that all makes sense. Thank you very much.
Thank you very much. I did get the derivative you show, but then divided through by 6 to get at the stationary points. Was that wrong?
I'm only looking for a critique please, not answers!
given: f(x)= −2x 3 +21x 2 − 60x+17
1. Is 1st deriv of this -x^2+7x-10 ?
2. stationary points x=2; x=5 ?
3. these are a local min and a local max?
Thanks
1. No. It is -6 x^2 +42 x-60.
2. See if the derivative is zero at both points. If so, the answer is yes.
3. Look at the second derivative. Evaluate it at x=2 and x=5. If the signs are + and -, the answer is yea. These cannot be absolute maxima and minima points because f(x) goes to + and - infinity.
Dividing by 6 is OK when setting the derivative equal to zero. When the derivative is zero, so is 1/6 the derivative. Note that 1/6 of the derivative is -(x-2)(x-5), so you can see right away where the stationary points are.
To find the stationary points, you need to set the derivative equal to zero and solve for x. In this case, the derivative is -6x^2 + 42x - 60. It seems like you got the correct derivative.
To find the solutions for x, you can either factor or use the quadratic formula. Since the equation is quadratic, factoring is the easier option. In this case, you can divide the derivative by 6 to get -(x^2 - 7x + 10), which can be factored as -(x-2)(x-5).
Setting (x-2)(x-5) equal to zero gives us two possible solutions for x: x = 2 and x = 5. So, the stationary points are x = 2 and x = 5.
Now let's move on to the next question. To determine whether these points are local minima or maxima, we need to look at the second derivative. The second derivative of f(x) is -12x + 42.
To evaluate the second derivative at x = 2, substitute x = 2 into the second derivative expression: -12(2) + 42 = 18. The sign of the second derivative at x = 2 is positive (+).
Now evaluate the second derivative at x = 5: -12(5) + 42 = -18. The sign of the second derivative at x = 5 is negative (-).
Since we have a change in sign from positive to negative, this indicates that we have a local maximum at x = 2 and a local minimum at x = 5.
However, it's important to note that these cannot be absolute maxima and minima points because the function f(x) goes to positive and negative infinity. Absolute maxima and minima occur when a function is bounded.
Hopefully, this explanation helps clarify the process for finding and analyzing the stationary points in this problem.