1. The problem asks which of the functions \(h(x)=\sqrt[3]{x+1}\) and \(f(x)=\sqrt[4]{x+1}\) are continuous at \(x=-2\). 2. To check continuity at \(x=-2\), we need to verify if the functions are defined at \(x=-2\) and if their limits as \(x\) approaches \(-2\) equal the function values. 3. For \(h(x)=\sqrt[3]{x+1}\), substitute \(x=-2\): $$h(-2)=\sqrt[3]{-2+1}=\sqrt[3]{-1}=-1$$ The cube root function is defined for all real numbers and is continuous everywhere. 4. For \(f(x)=\sqrt[4]{x+1}\), substitute \(x=-2\): $$f(-2)=\sqrt[4]{-2+1}=\sqrt[4]{-1}$$ The fourth root of a negative number is not a real number, so \(f(-2)\) is not defined in the real numbers. 5. Since \(h(x)\) is defined and continuous at \(x=-2\), and \(f(x)\) is not defined at \(x=-2\), only \(h(x)\) is continuous at \(x=-2\). **Final answer:** Only \(h(x)=\sqrt[3]{x+1}\) is continuous at \(x=-2\).