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$If f^(')(x)=-f(x) and f(1)=2, then f(3)=me^(n) for some integers m and n. What are m and n ? {:[m=],[n=]:}$

If $f^{\prime}(x)=-f(x)$ and $f(1)=2$ , then $f(3)=m e^{n}$ for some integers $m$ and $n$ . $\newline$ What are $m$ and $n$ ? $\newline$ $\begin{array}{l} m= \square \\ n= \square \end{array}$

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Transformations of quadratic functions

Full solution

Q. If

f^{\prime}(x)=-f(x)

and

f(1)=2

, then

f(3)=m e^{n}

for some integers

m

and

n

.

\newline

What are

m

and

n

?

\newline

\begin{array}{l} m= \square \\ n= \square \end{array}

Find $C$ value: Now, we use the initial condition $f(1) = 2$ to find the value of $C$ . Plugging $x = 1$ into $f(x) = Ce^{-x}$ , we get $f(1) = Ce^{-1} = 2$ .
Solve for C: Solve for C: $Ce^{-1} = 2$ implies $C = 2e$ .
Simplify function: Now we have the function $f(x) = 2e \cdot e^{-x}$ . Simplify this to get $f(x) = 2e^{1-x}$ .
Substitute $x=3$ : To find $f(3)$ , substitute $x = 3$ into $f(x) = 2e^{1-x}$ . So, $f(3) = 2e^{1-3}$ .
Calculate $f(3)$ : Calculate $f(3)$ : $f(3) = 2e^{-2}$ . This is the same as $f(3) = \frac{2}{e^2}$ .
Express in $me^{n}$ : Now, we need to express $\frac{2}{e^2}$ in the form of $me^{n}$ . Since $e^{-2}$ is the same as $\frac{1}{e^2}$ , we can write $f(3) = 2 \cdot e^{-2}$ .
Final result: Thus, we have $f(3) = 2 \cdot e^{(-2)}$ , which means $m = 2$ and $n = -2$ .

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h

is defined over the real numbers. This table gives a few values of

h

\newline

\begin{tabular}{lllll}

\newline

x

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1

-6

01

-6

001

-5

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\newline

h(x)

-0

25

-0

74

-0

98

-1

0

\newline

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-1

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