# Math Help - Examining a Curve - # 2

1. ## Examining a Curve - # 2

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

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

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

$f''(x) = 12x$

Find Critical Points:

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

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

$4x^{3} = 32$

$x^{3} = 8$

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

What is the maximum and minimum?

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

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

Where is the interval increasing or decreasing?

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

For $x<2$

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

For $2>x$

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

What is the inflection point?

$f''(x) = 12x$

$12x = 0$

$x = 0$

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

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

Inflection Point $(0,1)$

Where is the curve concave up, or concave down?

For $x<0$

$f''(x) = 12x$

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

For $x<0$

$f''(x) = 12x$

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

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

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

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

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

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

Find Critical Points:

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

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

$4x^{3} = 32$

$x^{3} = 8$

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

What is the maximum and minimum?

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

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

Where is the interval increasing or decreasing?

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

For $x<2$

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

For $2>x$

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

What is the inflection point?

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

$12x^{2} = 0$

$x^{2} = 0$

$x = 0$

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

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

Inflection Point $(0,1)$

Where is the curve concave up, or concave down?

For $x<0$

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

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

For $0 < x$

$f''(x) = 12x$

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