Hi, I don't understand this problem also.

Find the the relative maximmum and minimum of x^3 - 6x^2 + 15.
I'm supposed to do it using a graphing calculator but I still don't see any "lows" or "highs".

Relative max and mins will occur when the derivative is equal to zero. The derivative is 3x^2 - 12x. It equals 0 at x = 0 or x = 4. So they are your max and mins. The second derivative (deriv of the deriv) is 6x-12, which is negative at x=0 and positive at x=4. So since second derivative is negative at x = 0 it is a max, and since positive at x=4, it is a min. I hope derivatives were taught by this point in Pre calc, that is the only way I can explain it.

To find the relative maximum and minimum of the function x^3 - 6x^2 + 15, you can use a graphing calculator as instructed. If you're not able to observe the "lows" and "highs" visually from the graph, you can determine the critical points where the relative maxima and minima occur by evaluating the derivative of the function.

1. Start by finding the derivative of the function. The derivative of x^3 - 6x^2 + 15 with respect to x is given by:

f'(x) = 3x^2 - 12x.

2. Set the derivative equal to zero to find the critical points:

3x^2 - 12x = 0.

3. Factor out common terms:

3x(x - 4) = 0.

4. Solve for x by setting each factor equal to zero:

3x = 0 --> x = 0,
x - 4 = 0 --> x = 4.

5. Now, you have found the critical points at x = 0 and x = 4.

To determine whether these critical points are relative maxima or minima, you can use the second derivative test. The second derivative is obtained by taking the derivative of the first derivative:

f''(x) = 6x - 12.

6. Substitute the critical points, x = 0 and x = 4, into the second derivative to determine the sign of f''(x) at these points:

At x = 0: f''(0) = 6(0) - 12 = -12 (negative),
At x = 4: f''(4) = 6(4) - 12 = 12 (positive).

7. According to the second derivative test:
- If f''(x) is positive at a critical point, the function has a relative minimum at that point.
- If f''(x) is negative at a critical point, the function has a relative maximum at that point.

From the calculations, you can conclude that:
- x = 0 is a relative maximum,
- x = 4 is a relative minimum.

This procedure can help you find the relative maximum and minimum of the function x^3 - 6x^2 + 15 using the concepts of derivatives and the second derivative test.