Question

which function is represented by the graph below?

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

$f(x)=-2e^{x}$

$f(x)=e^{-2x}$

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

Explanation:

Step1: Analyze the domain of the functions

• For \(y = - 2\ln(x)\), the domain is \(x>0\). But the graph seems to have points for \(x <0\) (the left - hand side part). So this function is not correct.
• For \(y=-2e^{x}\), the domain is all real numbers. The range is \(y <0\). When \(x = 0\), \(y=-2e^{0}=-2\). As \(x

ightarrow-\infty\), \(y=-2e^{x}
ightarrow0\) (from the negative side).

• For \(y = e^{-2x}=(e^{-x})^{2}\), the domain is all real numbers and the range is \(y>0\). But the graph has \(y <0\) values, so this function is not correct.
• For \(y=-\ln(2x)\), the domain is \(x > 0\). But the graph seems to have points for \(x<0\) (the left - hand side part). So this function is not correct.

Step2: Check the behavior of \(y=-2e^{x}\)

• The general form of an exponential function is \(y = Ae^{kx}\), where \(A\) is the vertical stretch or compression factor and \(k\) affects the growth/decay rate.
• For \(y=-2e^{x}\), as \(x\) increases, \(y=-2e^{x}\) decreases (since the coefficient \(- 2<0\) and \(e^{x}\) is an increasing function). As \(x

ightarrow-\infty\), \(y=-2e^{x}
ightarrow0\) (because \(e^{x}
ightarrow0\) as \(x
ightarrow-\infty\))

Answer:

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