Section E Section E Checkpoint

Problem 1

A function is given by $f(x) = 2^x$ .

Write an exponential expression of the form $a \boldcdot b^x$ equal to $f(x+1)$ . Show your reasoning.
Show that $\frac{f(x+1)}{f(x)}$ is a constant.

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Solution

$2 \boldcdot 2^x$ because $f(x+1) = 2^{x+1}$ which is equal to $2 \boldcdot 2^x$
Sample response: $\frac{f(x+1)}{f(x)} = \frac{2^{x+1}}{2^x} = 2^{x+1-x} = 2^1$ which is 2.

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Sample Response

$2 \boldcdot 2^x$ because $f(x+1) = 2^{x+1}$ which is equal to $2 \boldcdot 2^x$
Sample response: $\frac{f(x+1)}{f(x)} = \frac{2^{x+1}}{2^x} = 2^{x+1-x} = 2^1$ which is 2.

Problem 2

The function $g$ is given by $g(x) = 3^x$ , and the function $h$ is given by $h(x) = 1,000 + 3x$ .

Show that $h(0) > g(0)$
For the same $x$ -value, will $g(x) > h(x)$ ? If so, give an example.

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Solution

$h(0) = 1000$ and $g(0) = 1$ , so $h(0) > g(0)$
Yes. Sample response: For any value greater than 7 this is true. For example, $g(7) = 2,187$ and $h(7) = 1,021$ .

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Sample Response

$h(0) = 1000$ and $g(0) = 1$ , so $h(0) > g(0)$
Yes. Sample response: For any value greater than 7 this is true. For example, $g(7) = 2,187$ and $h(7) = 1,021$ .