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

Consider function h(x): Let's first consider the function h(x) = ∛(x+1). To determine if h is continuous at x = -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 = -2 into h(x) to see if it is defined: h(-2) = ∛(-2+1) = ∛(-1). Since the cube root of any real number is defined, h(-2) is defined and equals -1.Check if h is continuous at x = -2: Now let's consider the function f(x) = ∜(x+1). To determine if f is continuous at x = -2, we need to check if the function is defined at that point. We substitute x = -2 into f(x) to see if it is defined: f(-2) = ∜(-2+1) = ∜(-1). The fourth root of a negative number is not defined in the real number system, so f(-2) is not defined.Substitute x = -2 into h(x): Since h(x) is defined at x = -2 and f(x) is not, only h(x) is continuous at x = -2. Therefore, the correct answer is (A) h only.

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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 $x=-2$ ? $\newline$ $\begin{array}{l} h(x)=\sqrt[3]{x+1} \\ f(x)=\sqrt[4]{x+1} \end{array}$ $\newline$ Choose $1$ answer: $\newline$ (A) $h$ only $\newline$ (B) $f$ only $\newline$ (C) Both $h$ and $f$ $\newline$ (D) Neither $h$ nor $f$

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Math Problems

Algebra 2

Domain and range of absolute value functions: equations

Full solution

Q. Which of the following functions are continuous at

x=-2

\newline

\begin{array}{l} h(x)=\sqrt[3]{x+1} \\ f(x)=\sqrt[4]{x+1} \end{array}

\newline

Choose

1

answer:

\newline

(A)

h

only

\newline

(B)

f

only

\newline

h

and

f

\newline

(D) Neither

h

nor

f

Consider function $h(x)$ : Let's first consider the function $h(x) = \sqrt[3]{x+1}$ . To determine if $h$ is continuous at $x = -2$ , 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 $x = -2$ into $h(x)$ to see if it is defined: $h(-2) = \sqrt[3]{-2+1} = \sqrt[3]{-1}$ . Since the cube root of any real number is defined, $h(-2)$ is defined and equals $-1$ .
Check if $h$ is continuous at $x = -2$ : Now let's consider the function $f(x) = \sqrt[4]{x+1}$ . To determine if $f$ is continuous at $x = -2$ , we need to check if the function is defined at that point. We substitute $x = -2$ into $f(x)$ to see if it is defined: $f(-2) = \sqrt[4]{-2+1} = \sqrt[4]{-1}$ . The fourth root of a negative number is not defined in the real number system, so $f(-2)$ is not defined.
Substitute $x = -2$ into $h(x)$ : Since $h(x)$ is defined at $x = -2$ and $f(x)$ is not, only $h(x)$ is continuous at $x = -2$ . Therefore, the correct answer is (A) $h$ only.