Question

For x < 0, find the maximum value of
(3x^2 +12)/x

Answer 1

The maximum value of the expression $\frac{(3x^2 + 12)}{x}$  for x < 0 is 0.

Explanation:

To find the maximum value of the expression $\frac{(3x^2 + 12)}{x}$ for x < 0, we can use calculus.

First, let's simplify the expression:

$\frac{(3x^2 + 12)}{x}$ = 3x + $\frac{12}{x}$

To find the maximum value, we need to find the critical points of the expression. Critical points occur where the derivative of the expression is equal to zero or undefined.

Taking the derivative of the expression with respect to x:

$\frac{d}{dx}$($\frac{(3x + 12)}{x}$) = 3 - $\frac{12}{x^{2} }$

Setting the derivative equal to zero:

3 - $\frac{12}{x^{2} }$ = 0

Rearranging the equation:

$\frac{12}{x^{2} }$ = 3.

Cross-multiplying:

12 = $3x^2$

Dividing by 3:

$4 = x^2$

Taking the square root:

x = ±2

Since we are looking for the maximum value for x < 0, we take x = -2.

Substituting x = -2 into the expression:

$\frac{(3(-2)^2 + 12)}{-2}$ = $\frac{(-12 + 12)}{-2}$ = 0

Therefore, the maximum value of the expression $\frac{(3x^2 + 12)}{x}$  for x < 0 is 0.

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