# Thread: Examining a Curve - # 2

1. ## Examining a Curve - # 2

All of this is wrong. Please look thru for mistakes.

$\displaystyle f(x) = x^{4} - 32x + 1$

$\displaystyle f'(x) = 4x^{3} - 32$

$\displaystyle f''(x) = 12x$

Find Critical Points:

$\displaystyle f'(x) = 4x^{3} - 32$

$\displaystyle 4x^{3} - 32 = 0$

$\displaystyle 4x^{3} = 32$

$\displaystyle x^{3} = 8$

$\displaystyle x = 2$ - $\displaystyle 3rd$ root of $\displaystyle 8$ is $\displaystyle 2$

What is the maximum and minimum?

$\displaystyle f(x) = x^{4} - 32x + 1$

$\displaystyle f(2) = (2)^{2} - 32(2) + 1 = -59$

Where is the interval increasing or decreasing?

$\displaystyle f'(x) = 4x^{3} - 32$

For $\displaystyle x<2$

$\displaystyle f'(1) = 4(1)^{3} - 32 = -21$ Negative so decreasing

For $\displaystyle 2>x$

$\displaystyle f'(3) = 4(x)^{3} - 32 = 4$ Positive so increasing

What is the inflection point?

$\displaystyle f''(x) = 12x$

$\displaystyle 12x = 0$

$\displaystyle x = 0$

$\displaystyle f(x) = x^{4} - 32x + 1$

$\displaystyle f(0) = (0)^{4} - 32(0) + 1 = 1$

Inflection Point $\displaystyle (0,1)$

Where is the curve concave up, or concave down?

For $\displaystyle x<0$

$\displaystyle f''(x) = 12x$

$\displaystyle f''(-1) = 12(-1) = -12$ Negative so concave up

For $\displaystyle x<0$

$\displaystyle f''(x) = 12x$

$\displaystyle f''(-1) = 12(-1) = -12$ Positive so concave down

2. ## Re: Examining a Curve - # 2

2nd derivative is corrected. Any hints on other errors?

$\displaystyle f(x) = x^{4} - 32x + 1$

$\displaystyle f'(x) = 4x^{3} - 32$

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

Find Critical Points:

$\displaystyle f'(x) = 4x^{3} - 32$

$\displaystyle 4x^{3} - 32 = 0$

$\displaystyle 4x^{3} = 32$

$\displaystyle x^{3} = 8$

$\displaystyle x = 2$ - $\displaystyle 3rd$ root of $\displaystyle 8$ is $\displaystyle 2$

What is the maximum and minimum?

$\displaystyle f(x) = x^{4} - 32x + 1$

$\displaystyle f(2) = (2)^{2} - 32(2) + 1 = -59$

Where is the interval increasing or decreasing?

$\displaystyle f'(x) = 4x^{3} - 32$

For $\displaystyle x<2$

$\displaystyle f'(1) = 4(1)^{3} - 32 = -21$ Negative so decreasing

For $\displaystyle 2>x$

$\displaystyle f'(3) = 4(x)^{3} - 32 = 4$ Positive so increasing

What is the inflection point?

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

$\displaystyle 12x^{2} = 0$

$\displaystyle x^{2} = 0$

$\displaystyle x = 0$

$\displaystyle f(x) = x^{4} - 32x + 1$

$\displaystyle f(0) = (0)^{4} - 32(0) + 1 = 1$

Inflection Point $\displaystyle (0,1)$

Where is the curve concave up, or concave down?

For $\displaystyle x<0$

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

$\displaystyle f''(-1) = 12(-1)^{2} = 12$ Negative so concave up

For $\displaystyle 0 < x$

$\displaystyle f''(x) = 12x$

$\displaystyle f''(1) = 12(1)^{2} = 12$ Positive so concave up