Let g(x)=(10 x)/(x^(3)+5x) when x!=0. g is continuous for all real numbers. Find g(0). Choose 1 answer: (A) 5 (B) 10 (C) 0 (D) 2

Simplify g(x): First, we need to simplify the function g(x) to see if it can be defined at x=0. g(x) = (10x) / (x^3 + 5x) We can factor out an x from the denominator. g(x) = (10x) / (x(x^2 + 5)) Since x is a common factor in the numerator and the denominator, we can cancel it out, as long as x is not equal to 0. g(x) = 10 / (x^2 + 5)Check continuity at x=0: Now, we need to check if g(x) is continuous at x=0 by substituting x with 0 in the simplified expression. g(0) = 10 / (0^2 + 5) g(0) = 10 / 5 g(0) = 2Conclusion: Since we have found a value for g(0) and the function does not have any discontinuities at x=0, g is continuous for all real numbers, including x=0.

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Let
g(x)=(10 x)/(x^(3)+5x) when
x!=0.

g is continuous for all real numbers.
Find
g(0).
Choose 1 answer:
(A) 5
(B) 10
(C) 0
(D) 2

Let $g(x)=\frac{10 x}{x^{3}+5 x}$ when $x \neq 0$ . $\newline$ $g$ is continuous for all real numbers. $\newline$ Find $g(0)$ . $\newline$ Choose $1$ answer: $\newline$ (A) $5$ $\newline$ (B) $10$ $\newline$ (C) $0$ $\newline$ (D) $2$

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Math Problems

Grade 8

Multiply and divide powers: variable bases

Full solution

Q. Let

g(x)=\frac{10 x}{x^{3}+5 x}

when

x \neq 0

\newline

g

is continuous for all real numbers.

\newline

Find

g(0)

\newline

Choose

1

answer:

\newline

(A)

5

\newline

(B)

10

\newline

(C)

0

\newline

(D)

2

Simplify $g(x)$ : First, we need to simplify the function $g(x)$ to see if it can be defined at $x=0$ .
$g(x) = \frac{10x}{x^3 + 5x}$
We can factor out an $x$ from the denominator.
$g(x) = \frac{10x}{x(x^2 + 5)}$
Since $x$ is a common factor in the numerator and the denominator, we can cancel it out, as long as $x \neq 0$ .
$g(x) = \frac{10}{x^2 + 5}$
Check continuity at $x=0$ : Now, we need to check if $g(x)$ is continuous at $x=0$ by substituting $x$ with $0$ in the simplified expression. $\newline$ $g(0) = \frac{10}{(0^2 + 5)}$ $\newline$ $g(0) = \frac{10}{5}$ $\newline$ $g(0) = 2$
Conclusion: Since we have found a value for $g(0)$ and the function does not have any discontinuities at $x=0$ , $g$ is continuous for all real numbers, including $x=0$ .