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Which of the following functions are continuous at x=-2 ? {:[h(x)=root(3)(x+1)],[f(x)=root(4)(x+1)]:} Choose 1 answer: (A) h only (B) f only (C) Both h and f (D) Neither h nor f

Which of the following functions are continuous at x=−2 x=-2 ?\newlineh(x)=x+13f(x)=x+14 \begin{array}{l} h(x)=\sqrt[3]{x+1} \\ f(x)=\sqrt[4]{x+1} \end{array} \newlineChoose 11 answer:\newline(A) h h only\newline(B) f f only\newline(C) Both h h and f f \newline(D) Neither h h nor f f

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Q. Which of the following functions are continuous at x=−2 x=-2 ?\newlineh(x)=x+13f(x)=x+14 \begin{array}{l} h(x)=\sqrt[3]{x+1} \\ f(x)=\sqrt[4]{x+1} \end{array} \newlineChoose 11 answer:\newline(A) h h only\newline(B) f f only\newline(C) Both h h and f f \newline(D) Neither h h nor f f
  1. Consider function h(x)h(x): Let's first consider the function h(x)=x+13h(x) = \sqrt[3]{x+1}. To determine if hh is continuous at x=−2x = -2, we need to check if the function is defined at that point and if there are no jumps or breaks in the graph at that point. We substitute x=−2x = -2 into h(x)h(x) to see if it is defined: h(−2)=−2+13=−13h(-2) = \sqrt[3]{-2+1} = \sqrt[3]{-1}. Since the cube root of any real number is defined, h(−2)h(-2) is defined and equals −1-1.
  2. Check if hh is continuous at x=−2x = -2: Now let's consider the function f(x)=x+14f(x) = \sqrt[4]{x+1}. To determine if ff is continuous at x=−2x = -2, we need to check if the function is defined at that point. We substitute x=−2x = -2 into f(x)f(x) to see if it is defined: f(−2)=−2+14=−14f(-2) = \sqrt[4]{-2+1} = \sqrt[4]{-1}. The fourth root of a negative number is not defined in the real number system, so f(−2)f(-2) is not defined.
  3. Substitute x=−2x = -2 into h(x)h(x): Since h(x)h(x) is defined at x=−2x = -2 and f(x)f(x) is not, only h(x)h(x) is continuous at x=−2x = -2. Therefore, the correct answer is (A) hh only.

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