Question

which function is represented by the graph below?
$f(x)=2 e^{x}$
$f(x)=ln x$
$f(x)=log (2 x)$
$f(x)=-e^{2 x}$

Explanation:

Step1: Analyze the domain of each function

• For \(y = 2e^{x}\), the domain is all real numbers \(x\in(-\infty,\infty)\).
• For \(y=\ln x\), the domain is \(x > 0\).
• For \(y=\log(2x)\), the domain is \(2x>0\) or \(x > 0\).
• For \(y=-e^{2x}\), the domain is all real numbers \(x\in(-\infty,\infty)\).

Step2: Analyze the general shape of the function

• The function \(y = 2e^{x}\) is an exponential growth function. Its graph is above the \(x\) - axis (\(y>0\)) for all \(x\) and passes through the point \((0,2)\) since when \(x = 0\), \(y=2e^{0}=2\times1 = 2\).
• The function \(y=\ln x\) has a vertical asymptote at \(x = 0\) and is a logarithmic growth function. When \(x = 1\), \(y=\ln(1)=0\).
• The function \(y=\log(2x)\) has a vertical asymptote at \(x = 0\). When \(x=\frac{1}{2}\), \(y=\log(2\times\frac{1}{2})=\log(1) = 0\).
• The function \(y=-e^{2x}\) is an exponential decay function. Its graph is below the \(x\) - axis (\(y<0\)) for all \(x\) since \(e^{2x}>0\) for all \(x\) and we have a negative sign.

Answer:

\(f(x)=2e^{x}\)