Questions: Let h(x)=sqrt(x^2+4) Where does h have critical points? Choose all answers that apply: A x=-2 (B) x=0 c x=2 D h has no critical points.

Transcript text: Let $h(x)=\sqrt{x^{2}+4}$ Where does $h$ have critical points? Choose all answers that apply: A $x=-2$ (B) $x=0$ c $x=2$ D $h$ has no critical points.

Solution

Solution Steps

Step 1: Find the Derivative of $ h(x) $

To find the critical points of the function $ h(x) = \sqrt{x^2 + 4} $, we first need to find its derivative. Using the chain rule, we have:

\[ h'(x) = \frac{d}{dx} \left( (x^2 + 4)^{1/2} \right) = \frac{1}{2}(x^2 + 4)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2 + 4}} \]

Step 2: Set the Derivative Equal to Zero

Critical points occur where the derivative is zero or undefined. We set the derivative equal to zero:

\[ \frac{x}{\sqrt{x^2 + 4}} = 0 \]

This equation is satisfied when the numerator is zero, i.e., $ x = 0 $.

Step 3: Check Where the Derivative is Undefined

The derivative $ \frac{x}{\sqrt{x^2 + 4}} $ is undefined when the denominator is zero. However, since $ \sqrt{x^2 + 4} $ is always positive for all real $ x $, the derivative is never undefined.