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How does $f(t) = 10^t$ change over the interval from $t = 9$ to $t = 10$ ? $\newline$ Choices: $\newline$ (A) $f(t)$ increases by $1,000\%$ $\newline$ (B) $f(t)$ decreases by a factor of $10$ $\newline$ (C) $f(t)$ increases by $10\%$ $\newline$ (D) $f(t)$ increases by $t = 9$ $0$

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Find instantaneous rates of change

Full solution

Q. How does

f(t) = 10^t

change over the interval from

t = 9

to

t = 10

?

\newline

Choices:

\newline

(A)

f(t)

increases by

1,000\%

\newline

(B)

f(t)

decreases by a factor of

10

\newline

(C)

f(t)

increases by

10\%

\newline

(D)

f(t)

increases by

t = 9

0

Calculate $f(t)$ : Calculate $f(t)$ when $t = 9$ : $f(9) = 10^9$ .
Calculate $f(t)$ : Calculate $f(t)$ when $t = 10$ : $f(10) = 10^{10}$ .
Find increase: Find the increase from $f(9)$ to $f(10)$ : $f(10) - f(9) = 10^{10} - 10^{9}$ .
Factor out base: Factor out the common base to simplify: $10^9(10 - 1) = 10^9 \times 9$ .
Calculate expression: Calculate the simplified expression: $10^9 \times 9 = 9 \times 10^9$ .
Compare increase: Compare the increase to the original value $f(9)$ : $\frac{9 \times 10^9}{10^9} = 9$ .

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