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A function $g(x)$ increases by a factor of $2$ over every unit interval in $x$ and $g(0) = 1$ . $\newline$ Which could be a function rule for $g(x)$ ? $\newline$ Choices: $\newline$ (A) $g(x) = 0.98^x$ $\newline$ (B) $g(x) = 2^x$ $\newline$ (C) $g(x) = 1 - \frac{x}{2}$ $\newline$ (D) $g(x) = -2x$

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Intermediate Value Theorem

Full solution

Q. A function

g(x)

increases by a factor of

2

over every unit interval in

x

and

g(0) = 1

.

\newline

Which could be a function rule for

g(x)

?

\newline

Choices:

\newline

(A)

g(x) = 0.98^x

\newline

(B)

g(x) = 2^x

\newline

(C)

g(x) = 1 - \frac{x}{2}

\newline

(D)

g(x) = -2x

Check $g(0)$ : Check $g(0)$ for each function to see if it equals $1$ . $\newline$ (A) $g(0) = 0.98^0 = 1$ (This matches $g(0) = 1$ , but we need to check if it doubles over each unit interval.) $\newline$ (B) $g(0) = 2^0 = 1$ (This matches $g(0) = 1$ , and we know that powers of $2$ double each time the exponent increases by $1$ .) $\newline$ (C) $g(0) = 1 - 0/2 = 1$ (This matches $g(0) = 1$ , but does it increase by a factor of $2$ over each unit interval?) $\newline$ (D) $g(0)$ $2$ (This does not match $g(0) = 1$ , so it's not the right function.)
Check doubling: Check if the functions double over each unit interval. $\newline$ (A) $g(1) = 0.98^1 \neq 2\cdot g(0)$ , so it doesn't double. $\newline$ (B) $g(1) = 2^1 = 2$ , which is $2\cdot g(0)$ , so it doubles. $\newline$ (C) $g(1) = 1 - \frac{1}{2} = \frac{1}{2}$ , which is not $2\cdot g(0)$ , so it doesn't double. $\newline$ (D) We already know (D) is incorrect because $g(0) \neq 1$ .
Choose correct function: Choose the function that both starts at $1$ and doubles over each unit interval. $\newline$ The correct function is (B) $g(x) = 2^x$ .

More problems from Intermediate Value Theorem

Question

Which of the following functions are continuous for all real numbers?

\newline

g(x)=\ln (x)

\newline

f(x)=\frac{1}{x}

\newline

1

\newline

g

\newline

f

\newline

g

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f(x)=\left\{\begin{array}{ll} \ln (-x)+3 & \text { for } x<-3 \\ \ln (-x+3) & \text { for }-3 \leq x<3 \end{array}\right.

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\newline

\mathrm{No}

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\newline

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\newline

f

x=2

\newline

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\newline

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\newline

h(x)=\left\{\begin{array}{ll} 2^{x-1} & \text { for } x<1 \\ 2^{1-x} & \text { for } x \geq 1 \end{array}\right.

\newline

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\newline

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\newline

g(x)=\left\{\begin{array}{ll} (x-2)^{2} & \text { for } x \leq 2 \\ 2-x^{2} & \text { for } x>2 \end{array}\right.

\newline

g

x=2

\newline

1

\newline

\newline

\mathrm{No}

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Question

\newline

g(x)=\left\{\begin{array}{ll} e^{x} & \text { for } x \leq-1 \\ -e^{-x} & \text { for } x>-1 \end{array}\right.

\newline

g

x=-1

\newline

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\newline

\newline

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\newline

h(x)=\left\{\begin{array}{ll} e^{2 x} & \text { for } x<0 \\ e^{5 x} & \text { for } x \geq 0 \end{array}\right.

\newline

h

x=0

\newline

1

\newline

\newline

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\newline

h(x)=\left\{\begin{array}{cl}\frac{6}{x+5} & \text { for }-5<x<-3 \\ x^{2}-6 & \text { for } x \geq-3\end{array}\right.

\newline

h

x=-3

\newline

1

\newline

\newline

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Question

\newline

g(x)=\left\{\begin{array}{ll}\sin (x) & \text { for } x<0 \\ x^{2} & \text { for } x \geq 0\end{array}\right.

\newline

g

x=0

\newline

1

\newline

\newline

Posted 3 months ago