How do you find the maximum value of #-3(sqrt(x-9) + 12)#?

Question

Luke Alvarez · Accepted Answer

maximum value: #-36#

You were probably intended to use derivatives, but here is a solution from a different approach.

Note that #sqrt(x-9) >= 0#

#hArr (sqrt(x-9)+12) >= 12#

#hArr -3(sqrt(x-9)+12) <= -36#

Christopher Allers · Answer

undefined To find the maximum value of $ -3(\sqrt{x-9} + 12) $:

1. Let $ y = -3(\sqrt{x-9} + 12) $.

2. Determine the domain of the function. Since the square root term $ \sqrt{x-9} $ requires the expression under the square root to be non-negative, we must have $ x - 9 \geq 0 $, which implies $ x \geq 9 $. So, the domain is $ x \geq 9 $.

3. To find the maximum value, we need to find the critical point where the derivative is equal to zero or undefined.

4. Take the derivative of $ y $ with respect to $ x $.

$$ y' = -3 \left( \frac{1}{2\sqrt{x-9}} \right) $$

5. Set the derivative equal to zero and solve for $ x $.

$$ -3 \left( \frac{1}{2\sqrt{x-9}} \right) = 0 $$

$$ \frac{1}{2\sqrt{x-9}} = 0 $$

The derivative is never undefined, so there are no critical points.

6. Since there are no critical points, the maximum value occurs at the endpoint of the domain.

7. Evaluate $ y $ at the endpoint of the domain.

$$ x = 9 $$

$$ y = -3(\sqrt{9-9} + 12) = -3(0 + 12) = -36 $$

So, the maximum value of the function $ -3(\sqrt{x-9} + 12) $ is $ -36 $, and it occurs at $ x = 9 $.