### Home > CALC > Chapter 7 > Lesson 7.1.4 > Problem 7-42

7-42.

*f*(*x*) = 2*x*^{3} − l5*x*^{2} + 24*x* + 19 for {*x*: 0 ≤ *x* ≤ 5}. Use the second derivative to justify that your value is a minimum. Homework Help ✎

Find *f* '(*x*) and *f* ''(*x*).

Identify candidates for minimum values by finding where *f* '(*x*) = 0 and where *f* '(*x*) = DNE. Note: this includes the endpoints.

Evaluate all candidates (except for the endpoints) in the 2nd derivative.

Recall that a LOCAL minimum will exist where *f* '(candidate) = 0 or DNE and *f* ''(candidate) > 0.

Note: endpoints can be global (not local) minima.

To find the GLOBAL minimum, evaluate and compare *f* (local minimum) with *f* (endpoint) and *f* (endpoint *b*).

The lowest value is the global minimum.