Question

the movement of the progress bar may be uneven because questions can be worth more or less (including
which function is represented by the graph below?
graph of a curve
$f(x) = e \cdot x$ $f(x) = e^x$ $f(x) = \log x$ $f(x) = \ln x$

Explanation:

Step1: Analyze the function \(f(x)=e\cdot x\)

This is a linear function. Its graph is a straight line. But the given graph is a curve, so \(f(x) = e\cdot x\) is not the correct function.

Step2: Analyze the function \(f(x)=\log x\)

The domain of \(y = \log x\) is \(x>0\). When \(x = 1\), \(y=\log(1)=0\). But in the given graph, when \(x = 0\), the function value is not \(0\) and the general shape of the logarithmic function \(y=\log x\) (it grows more slowly compared to the exponential function for \(x>1\)) does not match the given graph.

Step3: Analyze the function \(f(x)=\ln x\)

The domain of \(y=\ln x\) is \(x > 0\). When \(x = 1\), \(y=\ln(1)=0\). The graph of \(y = \ln x\) is different from the given graph (it has a different rate of growth and passes through \((1,0)\)).

Step4: Analyze the function \(f(x)=e^{x}\)

The function \(y = e^{x}\) has the property that when \(x = 0\), \(y=e^{0}=1\). As \(x\) increases, \(y = e^{x}\) grows exponentially. For example, when \(x = 1\), \(y=e^{1}\approx2.718\); when \(x = 2\), \(y=e^{2}\approx7.389\). The general shape of the exponential function \(y = e^{x}\) (increasing, passes through \((0,1)\) and has rapid growth for \(x>0\)) matches the given graph.

Answer:

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