1. **Problem Statement:** We want to analyze the concavity and inflection points of the curve given by the function $$y=\frac{1}{x^2}+4$$. 2. **Formula and Rules:** - The concavity of a function is determined by the sign of its second derivative $y''$. - If $y''>0$, the curve is concave up. - If $y''<0$, the curve is concave down. - Inflection points occur where $y''=0$ or $y''$ is undefined and the concavity changes. 3. **First Derivative:** Given $$y=\frac{1}{x^2}+4 = x^{-2} + 4,$$ we differentiate: $$y' = -2x^{-3} = -\frac{2}{x^3}.$$ 4. **Second Derivative:** Differentiate $y'$: $$y'' = 6x^{-4} = \frac{6}{x^4}.$$ 5. **Analyze Concavity:** Since $x^4 > 0$ for all $x \neq 0$, we have: $$y'' = \frac{6}{x^4} > 0 \quad \text{for all } x \neq 0.$$ This means the curve is concave up everywhere except at $x=0$ where the function is undefined. 6. **Inflection Points:** Because $y''$ never equals zero and does not change sign, there are no inflection points. **Final answer:** The curve $$y=\frac{1}{x^2}+4$$ is concave up for all $x \neq 0$ and has no inflection points.