Find Intervals of Increase and Decrease: Analyzing y = √5x² + 4

To solve the problem of finding the intervals of increase and decrease for the function $y = \sqrt{5}x^2 + 4$, we will follow these steps:

• Step 1: Identify the function $y = \sqrt{5}x^2 + 4$. Here $a = \sqrt{5}$, so the parabola opens upwards.
• Step 2: Compute the derivative of the function, $y' = \frac{dy}{dx}$, which will help us find the critical points where the slope changes.

Calculating the derivative, we have:

$y = \sqrt{5}x^2 + 4$

$y' = \frac{d}{dx} (\sqrt{5}x^2 + 4) = 2\sqrt{5}x$

Step 3: Find the critical points by setting the derivative equal to zero:

$2\sqrt{5}x = 0$

Solving for $x$ gives $x = 0$.

Step 4: Analyze the sign of $y' = 2\sqrt{5}x$ to determine where the function is increasing or decreasing:

• For $x < 0$, $2\sqrt{5}x < 0$, so the function is decreasing.
• For $x > 0$, $2\sqrt{5}x > 0$, so the function is increasing.

Hence, the function decreases on the interval $x < 0$ and increases on the interval $x > 0$.

Therefore, the solution to the problem is that the function is decreasing for $x > 0$ and increasing for $x < 0$.

$\searrow:x > 0$ and $\nearrow:x < 0$

According to the provided answer choices, the correct choice that matches the intervals we found is choice 3: $\searrow:x > 0$ and $\nearrow:x < 0$.