A function $g(x)$ increases by $2$ over every unit interval in $x$ and $g(0) = 0$ . $\newline$ Which could be a function rule for $g(x)$ ? $\newline$ Choices: $\newline$ (A) $g(x) = -x + 2$ $\newline$ (B) $g(x) = -2x$ $\newline$ (C) $g(x) = 2x$ $\newline$ (D) $g(x) = -\frac{x}{2}$

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Intermediate Value Theorem

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Q. A function

g(x)

increases by

2

over every unit interval in

x

and

g(0) = 0

\newline

Which could be a function rule for

g(x)

\newline

Choices:

\newline

(A)

g(x) = -x + 2

\newline

(B)

g(x) = -2x

\newline

(C)

g(x) = 2x

\newline

(D)

g(x) = -\frac{x}{2}

Eliminate Incorrect Choices: Since $g(0) = 0$ , we can eliminate any choice that doesn't give us $0$ when $x$ is $0$ . $\newline$ Let's check each option.
Check Option (A): For (A) $g(x) = -x + 2$ , if we put $x = 0$ , we get $g(0) = -0 + 2 = 2$ , which doesn't match $g(0) = 0$ . $\newline$ So, (A) is not the right function.
Check Option (B): For (B) $g(x) = -2x$ , if we put $x = 0$ , we get $g(0) = -2\times 0 = 0$ , which matches $g(0) = 0$ . But, we need to check if it increases by $2$ over every unit interval.
Check Option (C): If we increase $x$ by $1$ , so $x = 1$ , then $g(1) = -2 \times 1 = -2$ , which means the function decreases by $2$ , not increases. $\newline$ So, (B) is not the right function either.
Check Option (D): For (C) $g(x) = 2x$ , if we put $x = 0$ , we get $g(0) = 2\times 0 = 0$ , which matches $g(0) = 0$ . Now let's check the increase over each unit interval.
Check Option (D): For (C) $g(x) = 2x$ , if we put $x = 0$ , we get $g(0) = 2\times0 = 0$ , which matches $g(0) = 0$ . Now let's check the increase over each unit interval.If we increase $x$ by $1$ , so $x = 1$ , then $g(1) = 2\times1 = 2$ , which means the function increases by $2$ , which is what we want. So, (C) seems to be the right function.
Check Option (D): For (C) $g(x) = 2x$ , if we put $x = 0$ , we get $g(0) = 2\times 0 = 0$ , which matches $g(0) = 0$ . Now let's check the increase over each unit interval.If we increase $x$ by $1$ , so $x = 1$ , then $g(1) = 2\times 1 = 2$ , which means the function increases by $2$ , which is what we want. So, (C) seems to be the right function.For (D) $g(x) = -\frac{x}{2}$ , if we put $x = 0$ , we get $x = 0$ $1$ , which matches $g(0) = 0$ . But we need to check the increase over each unit interval.
Check Option (D): For (C) $g(x) = 2x$ , if we put $x = 0$ , we get $g(0) = 2\times 0 = 0$ , which matches $g(0) = 0$ . Now let's check the increase over each unit interval. If we increase $x$ by $1$ , so $x = 1$ , then $g(1) = 2\times 1 = 2$ , which means the function increases by $2$ , which is what we want. So, (C) seems to be the right function. For (D) $g(x) = -\frac{x}{2}$ , if we put $x = 0$ , we get $x = 0$ $1$ , which matches $g(0) = 0$ . But we need to check the increase over each unit interval. If we increase $x$ by $1$ , so $x = 1$ , then $x = 0$ $6$ , which means the function decreases, not increases. So, (D) is not the right function.