How does f(t) = 2^t change over the interval from t = 5 to t = 6? Choices: (A)f(t) increases by a factor of 2 (B)f(t) decreases by a factor of 2 (C)f(t) increases by 200% (D)f(t) increases by 2

Calculate f(t) at t=5: Calculate f(t) when t = 5. f(5) = 2^5 f(5) = 32Calculate f(t) at t=6: Calculate f(t) when t = 6. f(6) = 2^6 f(6) = 64Find change in f(t): Find the change in f(t) from t = 5 to t = 6. Change = f(6) - f(5) Change = 64 - 32 Change = 32Determine factor of increase: Determine the factor by which f(t) increases. Factor = f(6) / f(5) Factor = 64 / 32 Factor = 2

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How does $f(t) = 2^t$ change over the interval from $t = 5$ to $t = 6$ ? $\newline$ Choices: $\newline$ (A) $f(t)$ increases by a factor of $2$ $\newline$ (B) $f(t)$ decreases by a factor of $2$ $\newline$ (C) $f(t)$ increases by $200\%$ $\newline$ (D) $f(t)$ increases by $2$

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Math Problems

Calculus

Find derivatives of logarithmic functions

Full solution

Q. How does

f(t) = 2^t

change over the interval from

t = 5

t = 6

\newline

Choices:

\newline

(A)

f(t)

increases by a factor of

2

\newline

(B)

f(t)

decreases by a factor of

2

\newline

(C)

f(t)

increases by

200\%

\newline

(D)

f(t)

increases by

2

Calculate $f(t)$ at $t=5$ : Calculate $f(t)$ when $t = 5$ . $\newline$ $f(5) = 2^5$ $\newline$ $f(5) = 32$
Calculate $f(t)$ at $t=6$ : Calculate $f(t)$ when $t = 6$ . $\newline$ $f(6) = 2^6$ $\newline$ $f(6) = 64$
Find change in $f(t)$ : Find the change in $f(t)$ from $t = 5$ to $t = 6$ . $\newline$ Change = $f(6) - f(5)$ $\newline$ Change = $64 - 32$ $\newline$ Change = $32$
Determine factor of increase: Determine the factor by which $f(t)$ increases. $\newline$ Factor = $\frac{f(6)}{f(5)}$ $\newline$ Factor = $\frac{64}{32}$ $\newline$ Factor = $2$