Question:

Step 1: Substitute to simplify the function using a new variable. Let $t = x^2$. Since $x^2 \geq 0$ for all real $x$, we have $t \geq 0$. The function becomes: $$h(t) = t + \frac{1}{t+1} \text{ where } t \geq 0$$ Step 2: Find the critical points by taking the derivative. To find the extrema of $h(t)$, we compute the derivative: $$h'(t) = 1 - \frac{1}{(t+1)^2}$$ Step 3: Set the derivative equal to zero and solve for critical points. Setting $h'(t) = 0$: $$1 - \frac{1}{(t+1)^2} = 0$$ $$(t+1)^2 = 1$$ Since $t \geq 0$, we have $t + 1 \geq 1 > 0$, so: $$t + 1 = 1$$ $$t = 0$$ Step 4: Evaluate the function at the critical point. At $t = 0$: $$h(0) = 0 + \frac{1}{0+1} = 1$$ Step 5: Analyze the behavior of the function for $t > 0$. For $t > 0$, we have $(t+1)^2 > 1$, which means $\frac{1}{(t+1)^2} < 1$, so: $$h'(t) = 1 - \frac{1}{(t+1)^2} > 0$$ This shows that $h(t)$ is strictly increasing for all $t > 0$. Step 6: Determine the behavior as $t \to \infty$. As $t \to \infty$: $$h(t) = t + \frac{1}{t+1} \to \infty + 0 = \infty$$ Step 7: Conclude the range. Since $h(t)$ has a minimum value of $1$ at $t = 0$ and increases without bound as $t \to \infty$, the range of the function is: $$\boxed{[1, \infty)}$$ The answer is **Option 1: $[1, \infty)$**