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=

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:

.

This polynomial has roots such that , given by the following formulas:

In this question:

Finding the critical points:

Simplifying by 3

So

Second derivative test:

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

The second derivative is:

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

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

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

answ9x x 3=

step-by-step explanation: