Max and Min Problem - # 2

[Jason76]

\(\displaystyle f(x) = 6x^{3} - 9x^{2} - 216x + 5\)

\(\displaystyle f'(x) = 18x^{2} - 18x - 216\)

\(\displaystyle f'(x) = 18(x^{2} - x - 12)\)

\(\displaystyle 18(x^{2} - x - 12) = 0\)

\(\displaystyle (x + 3)(x - 4) = 0\)

Critical Points:

\(\displaystyle x = -3\)

\(\displaystyle x = 4\)

2nd Derivative Test:

\(\displaystyle f''(x) = \dfrac{d^{2}}{dx} 6x^{3} - 9x^{2} - 216x + 5\)

\(\displaystyle f''(x) = [(x^{2} - x - 12) ][0] + [18][(2x - 1)]\)

\(\displaystyle f''(x) = [18][(2x - 1)]\)

\(\displaystyle f''(x) = 32x - 18\)

for \(\displaystyle x = -3\)

\(\displaystyle f''(x) = 32(-3) - 18 = -118\) - Maximum is \(\displaystyle x = -3\)

for \(\displaystyle x = 4\)

\(\displaystyle f''(x) = 32(4) - 18 = 110 \) - Minimum is for \(\displaystyle x = 4\)

:?: Online homework says this is wrong.

 

[stapel]

What are the instructions for this exercise? What are you supposed to be doing with the posted function?

\(\displaystyle f(x) = 6x^{3} - 9x^{2} - 216x + 5\)

\(\displaystyle f'(x) = 18x^{2} - 18x - 216\)

2nd Derivative Test:

\(\displaystyle f''(x) = \dfrac{d^{2}}{dx} 6x^{3} - 9x^{2} - 216x + 5\)

The second derivative of f(x) is the first derivative of f'(x). So try differentiating f'(x). This may simplify your calculations so you get the correct values for the coefficients.

 

[HallsofIvy]

\(\displaystyle f(x) = 6x^{3} - 9x^{2} - 216x + 5\)

\(\displaystyle f'(x) = 18x^{2} - 18x - 216\)

\(\displaystyle f'(x) = 18(x^{2} - x - 12)\)

\(\displaystyle 18(x^{2} - x - 12) = 0\)

\(\displaystyle (x + 3)(x - 4) = 0\)

Critical Points:

\(\displaystyle x = -3\)

\(\displaystyle x = 4\)

2nd Derivative Test:

\(\displaystyle f''(x) = \dfrac{d^{2}}{dx} 6x^{3} - 9x^{2} - 216x + 5\)

\(\displaystyle f''(x) = [(x^{2} - x - 12) ][0] + [18][(2x - 1)]\)

\(\displaystyle f''(x) = [18][(2x - 1)]\)

You had, before, that \(\displaystyle f'(x)= 18(x^2- x- 12)\) so that you could easily get \(\displaystyle f''= 18(2x- 1)\) from that.

\(\displaystyle f''(x) = 32x - 18\)

But here is your error- 2(18)= 36, not 32. In any case you are only concerned with the sign. Since "18" is positive, it is sufficient to look at "2x- 1" for x= -3 and 4.

for \(\displaystyle x = -3\)

\(\displaystyle f''(x) = 32(-3) - 18 = -118\) - Maximum is \(\displaystyle x = -3\)

for \(\displaystyle x = 4\)

\(\displaystyle f''(x) = 32(4) - 18 = 110 \) - Minimum is for \(\displaystyle x = 4\)

:?: Online homework says this is wrong.

Since you had already determined that f'= 18(x+ 3)(x- 4). If x< -3, both x+3 and x- 4 are negative so their product is positive so f' is positive. For x between -3 and 4, x+ 3 is positive while x- 4 is still negative so their product is negative so f' is negative. For x> 4, both x+ 3 and x- 4 are positive so their product is positive so f' is positive. That is, f is increasing up to x=-3, then decreasing to x= 4, then increasing again.

 

[lookagain]

> > Critical Points: < <

\(\displaystyle x = -3\)

\(\displaystyle x = 4\)

Jason76, those are critical numbers, not critical points.

If the problem requires critical points, you have to evaluate the function at those critical numbers

to get the respective y-values to complete the critical points, which are sets of coordinates.

 

[Deleted member 4993]

\(\displaystyle f(x) = 6x^{3} - 9x^{2} - 216x + 5\)




:?: Online homework says this is wrong.

You did not specify "exactly" what you were to calculate?

 

[Jason76]

You did not specify "exactly" what you were to calculate?

We need to find the respective x values which correspond to a max and min.

 

[Deleted member 4993]

We need to find the respective x values which correspond to a max and min.

There was no restricted domain?
Then "local" maximum at x = -4 and "local" minimum at x = 4 is correct.