Question

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

Explanation:

Step1: Analyze the domain of the function

For \(y = e^{-2x}\), the domain is all real numbers. For \(y=-2e^{x}\), the domain is all real numbers. For \(y =-\ln(2x)\), the domain is \(x>0\). For \(y=-2\ln(x)\), the domain is \(x > 0\). From the graph, we can see that the function is defined for \(x>0\).

Step2: Check the value of the function at a particular point

Let's take \(x = 1\).

• For \(y=-\ln(2x)\), when \(x = 1\), \(y=-\ln(2)\approx - 0.69\)
• For \(y = e^{-2x}\), when \(x = 1\), \(y=e^{-2}=\frac{1}{e^{2}}\approx0.14\)
• For \(y=-2\ln(x)\), when \(x = 1\), \(y=-2\ln(1)=0\)
• For \(y=-2e^{x}\), when \(x = 1\), \(y=-2e\approx - 5.44\)

Step3: Analyze the end - behavior

As \(x
ightarrow+\infty\):

• For \(y =-\ln(2x)\), \(y

ightarrow-\infty\)

• For \(y = e^{-2x}\), \(y

ightarrow0\)

• For \(y=-2\ln(x)\), \(y

ightarrow-\infty\)

• For \(y=-2e^{x}\), \(y

ightarrow-\infty\)

As \(x
ightarrow0^{+}\):

• For \(y =-\ln(2x)\), \(y

ightarrow+\infty\)

• For \(y=-2\ln(x)\), \(y

ightarrow+\infty\)

We can also use the fact that the general form of a logarithmic function \(y = a\ln(x)+b\) and an exponential function \(y = Ae^{kx}+C\). The shape of the given graph is more similar to the shape of a logarithmic function rather than an exponential function.

Answer:

\(f(x)=-2\ln(x)\)