Question

For the function, find all critical numbers and then use the second-derivative test to determine whether the function has a relative maximum or minimum at each critical number. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
f(x) = x3 − 12x2 + 21x − 8relative maxima x=relative minima x=

Answer 1

Answer:

relative maximum: x = 1

relative minimum: x = 7

Step-by-step explanation:

Critical points:

Values of x for which f'(x) = 0.

Second derivative test:

For a critical point, if f''(x) > 0, the critical point is a relative minimum.

Otherwise, if f''(x) < 0, the critical point is a relative maximum.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this question:

$f(x) = x^{3} - 12x^{2} + 21x - 8$

Finding the critical points:

$f'(x) = 3x^{2} - 24x + 21$

$3x^{2} - 24x + 21 = 0$

Simplifying by 3

$x^{2} - 8x + 7 = 0$

So $a = 1, b = -8, c = 7$

$\bigtriangleup = (-8)^{2} - 4*1*7 = 36$

$x_{1} = \frac{-(-8) + \sqrt{36}}{2} = 7$

$x_{2} = \frac{-(-8) - \sqrt{36}}{2} = 1$

Second derivative test:

The critical points are x = 1 and x = 7.

The second derivative is:

$f''(x) = 6x - 24$

$f''(1) = 6*1 - 24 = -18$

Since f''(1) < 0, at x = 1 there is a relative maximum.

$f''(7) = 6*7 - 24 = 18$

Since f''(x) > 0, at x = 7 there is a relative minumum.