1. **State the problem:** We are given the function $f(x) = x^2 + 1$ and asked to analyze it using calculus. 2. **Recall the formula for the derivative:** The derivative of a function $f(x)$, denoted $f'(x)$, gives the rate of change or slope of the function at any point $x$. For power functions, the rule is $\frac{d}{dx} x^n = n x^{n-1}$. 3. **Find the derivative:** $$ f'(x) = \frac{d}{dx} (x^2 + 1) = \frac{d}{dx} x^2 + \frac{d}{dx} 1 = 2x + 0 = 2x $$ 4. **Interpret the derivative:** The slope of the function at any point $x$ is $2x$. This means the slope is zero at $x=0$, positive for $x>0$, and negative for $x<0$. 5. **Find critical points:** Set $f'(x) = 0$ to find critical points. $$ 2x = 0 \implies x = 0 $$ 6. **Determine the nature of the critical point:** Use the second derivative test. The second derivative is $$ f''(x) = \frac{d}{dx} (2x) = 2 $$ Since $f''(0) = 2 > 0$, the function has a local minimum at $x=0$. 7. **Find the minimum value:** $$ f(0) = 0^2 + 1 = 1 $$ **Final answer:** The function $f(x) = x^2 + 1$ has a local minimum at $x=0$ with minimum value $1$. The derivative is $f'(x) = 2x$ and the function is increasing for $x>0$ and decreasing for $x<0$.