### Home > APCALC > Chapter 7 > Lesson 7.1.4 > Problem7-39

7-39.

Without a calculator, determine the minimum value of $f(x) = 2x^3 - 15x^2 + 24x + 19$ for $0 ≤ x ≤ 5$. Use the second derivative to justify that your value is a minimum.

Find $f^\prime(x)$ and $f ^{\prime\prime}(x)$.

Identify candidates for minimum values by finding where $f^\prime(x)=0$ and where $f^\prime(x)=$ 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$f^\prime(\text{candidate}) = 0$ or DNE and $f ^{\prime\prime}(\text{candidate}) > 0$.
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.