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Math

Math Snap

PROBLEM

\begin{align}

1. & \quad f(x) = 3\left(e^{x} + e^{-x}\right) \\

3. & \quad f(w) = \frac{e^{w} + 2}{e^{w}} \\

5. & \quad f(x) = 2 e^{3x-1} \end{align
}

STEP 1

1. We are given three separate functions, each defined in terms of the exponential function ex e^x .
2. We need to analyze each function individually.
3. The goal is to understand the structure and behavior of each function.

STEP 2

1. Analyze the function f(x)=3(ex+ex) f(x) = 3(e^x + e^{-x}) .
2. Analyze the function f(w)=ew+2ew f(w) = \frac{e^w + 2}{e^w} .
3. Analyze the function f(x)=2e3x1 f(x) = 2e^{3x-1} .

STEP 3

Consider the function f(x)=3(ex+ex) f(x) = 3(e^x + e^{-x}) .
- Recognize that this is a combination of exponential growth and decay terms.
- The term ex e^x represents exponential growth, while ex e^{-x} represents exponential decay.
- The function can be rewritten using hyperbolic cosine: f(x)=32cosh(x) f(x) = 3 \cdot 2 \cosh(x) , where cosh(x)=ex+ex2 \cosh(x) = \frac{e^x + e^{-x}}{2} .
- This function is symmetric about the y-axis, as f(x)=f(x) f(-x) = f(x) .

STEP 4

Consider the function f(w)=ew+2ew f(w) = \frac{e^w + 2}{e^w} .
- Simplify the expression by dividing each term in the numerator by ew e^w .
- This results in: f(w)=1+2ew f(w) = 1 + \frac{2}{e^w} .
- As w w \to \infty , ew e^w \to \infty and 2ew0 \frac{2}{e^w} \to 0 , so f(w)1 f(w) \to 1 .
- As w w \to -\infty , ew0 e^w \to 0 and 2ew \frac{2}{e^w} \to \infty , so f(w) f(w) \to \infty .

SOLUTION

Consider the function f(x)=2e3x1 f(x) = 2e^{3x-1} .
- Recognize that this is an exponential function with a transformation.
- The exponent 3x1 3x - 1 indicates a horizontal stretch/compression and shift.
- Rewrite the exponent as e3xe1 e^{3x} \cdot e^{-1} , which simplifies to 2ee3x \frac{2}{e} \cdot e^{3x} .
- This function represents exponential growth, with a growth rate determined by the factor 3 3 .
Each function has been analyzed in terms of its structure and behavior.

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