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$If f^(')(x)=f(x) and f(3)=e^(2), then f(4)=me^(n) for some integers m and n. What are m and n ? {:[m=],[n=]:}$

If $f^{\prime}(x)=f(x)$ and $f(3)=e^{2}$ , then $f(4)=m e^{n}$ 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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Find the vertex of the transformed function

Full solution

Q. If

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

and

f(3)=e^{2}

, then

f(4)=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}

Given Differential Equation: We are given that $f'(x) = f(x)$ . This is the differential equation for the exponential function. The general solution to this differential equation is $f(x) = Ce^{x}$ , where $C$ is a constant.
Finding Constant C: We are also given that $f(3) = e^2$ . We can use this information to find the value of the constant $C$ . Substituting $x = 3$ into the general solution, we get $f(3) = Ce^{3}$ . Since $f(3) = e^2$ , we can set $Ce^{3} = e^2$ and solve for $C$ .
Specific Solution: Solving $Ce^{3} = e^{2}$ for $C$ gives us $C = e^{2-3} = e^{-1}$ . So the specific solution to the differential equation that satisfies the given condition is $f(x) = e^{-1}e^{x} = e^{x-1}$ .
Finding $f(4)$ : Now we want to find $f(4)$ . Substituting $x = 4$ into the specific solution $f(x) = e^{(x-1)}$ , we get $f(4) = e^{(4-1)} = e^{3}$ .
Final Expression: We can express $e^{3}$ as $e^{2} \times e^{1}$ . Since $e^{2}$ is already given, we can write $e^{3}$ as $e^{2} \times e$ , which is the same as $1 \times e^{2} \times e$ . Therefore, $m = 1$ and $n = 2 + 1 = 3$ .

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