Question

Given the following function:

F (X) = -93 - 31X2 - 19X

1. What is the value of the critical point for this function?

2. Is this critical point a maximum, minimum or is it indeterminate?

Answer 1

(1) To obtain critical value(s), the equation is differentiated and equated to zero. Given $F (X) = -93 - 31X^2 - 19X$. The first derivative is given as

$\frac{dy}{dx}(−31x^2−19x−93)$

$=−31\frac{d}{dx}(x^2)−19\frac{d}{dx}(x)+\frac{d}{dx}(−93)$

$=−62x−19+0=−62x−19$

Equating to zero, we have

$0=−62x−19$

$19=-62x$

$x=\frac{-19}{62}$

Then $x$ is substituted into $F(X)$ to get $y$ coordinate. Thus,

$F (\frac{-19}{62}) = -93 - 31(\frac{-19}{62})^2 - 19\frac{-19}{62}$

$=-93-31(\frac{361}{3844})+\frac{361}{62}$

$y=-93-\frac{11191}{3844}+\frac{361}{62}$

$y=-93+\frac{11191}{3844}=-90.0089$.

Therefore, the critical point of the equation is at $x=\frac{-19}{62}$ and the value $y=-90.0089$

(2). To determine whether the critical point is maximum, minimum or indeterminate, the second derivative is required. The second derivative is given by

$\frac{d}{dx}(−62x−19)$

$=−62\frac{d}{dx}(x)+\frac{d}{dx}(−19)$

$=0−62 =−62$

Since the value of the second derivative is negative, the critical point is a maximum.