Question

c)

options:
\\( f(x) = \left(\frac{1}{3}\
ight)^x \\)
\\( f(x) = \log_3(x) \\)
\\( f(x) = \log_{\frac{1}{3}}(x) \\)
\\( f(x) = 3^x \\)

Explanation:

Step1: Analyze the graph's shape

The graph is an exponential curve (since it has the shape of an exponential function, not logarithmic which has a vertical asymptote and is defined for \(x>0\) only here the graph is defined for all \(x\) and passes through \((0,1)\) as \(3^0 = 1\), \((\frac{1}{3})^0=1\) but the growth is increasing for \(f(x) = 3^x\) (since base \(3>1\)) and decreasing for \(f(x)=(\frac{1}{3})^x\) (base \(0<\frac{1}{3}<1\)). The graph here is increasing (rising to the right) so it should be an exponential function with base \(>1\).

Step2: Check the options

• Option 1: \(f(x)=(\frac{1}{3})^x\) is a decreasing exponential (base \(<1\)), so its graph would be decreasing as \(x\) increases, but our graph is increasing. Eliminate.
• Option 2: \(f(x)=\log_3(x)\) is a logarithmic function, defined only for \(x > 0\), but our graph is defined for all \(x\) (including \(x\leq0\)), so eliminate.
• Option 3: \(f(x)=\log_{\frac{1}{3}}(x)\) is also a logarithmic function, defined for \(x>0\), eliminate.
• Option 4: \(f(x) = 3^x\) is an exponential function with base \(3>1\), so it is increasing, passes through \((0,1)\) (since \(3^0 = 1\)), and as \(x\to-\infty\), \(3^x\to0\) (matches the horizontal asymptote on the left), and as \(x\) increases, it grows rapidly (matches the graph).

Answer:

\(f(x) = 3^x\) (the fourth option: \(f(x) = 3^x\))