Q. Which is the graph of \( f(x) = 2(3)^x \)?

Answer

\( f(x) = 2\cdot 3^{x} \)

Quick explanation:

  • \( f(0) = 2 \), \( f(1) = 6 \), \( f(-1) = \tfrac{2}{3} \).
  • Horizontal asymptote: \( y = 0 \).
  • Strictly increasing exponential growth (base \( 3 > 1 \)), passes through \( (0,2) \) and rises rapidly to the right.

Final result: The graph is an exponential growth curve with y-intercept 2 and horizontal asymptote \( y = 0 \).

Detailed Explanation

Problem

Graph the function \( f(x) = 2 \cdot 3^{x} \).

Step-by-step explanation

  1. Recognize the type of function.

    The function is an exponential function of the form \( f(x) = a \cdot b^{x} \) with base \( b = 3 \) and vertical scale factor \( a = 2 \). This means it is exponential growth because \( b > 1 \).

  2. Domain.

    Exponential functions are defined for all real inputs. Therefore the domain is
    \[ \text{Domain} = (-\infty, \infty). \]

  3. Range.

    For any real \( x \), \( 3^{x} > 0 \). Multiplying by 2 keeps the values positive. Thus
    \[ \text{Range} = (0, \infty). \]

  4. Horizontal asymptote.

    As \( x \) becomes very negative, \( 3^{x} \) approaches 0. Multiplying by 2 still approaches 0, so the horizontal asymptote is
    \[ y = 0. \]

  5. Intercepts.

    Compute the y-intercept by evaluating at \( x = 0 \):
    \[ f(0) = 2 \cdot 3^{0} = 2 \cdot 1 = 2. \]
    So the graph passes through \( (0,2) \).

    There is no x-intercept because \( f(x) > 0 \) for all \( x \), so the graph never crosses the x-axis.

  6. Monotonicity (increasing/decreasing).

    Differentiate to check the slope:
    \[ f'(x) = 2 \cdot 3^{x} \cdot \ln 3. \]
    Since \( 2 > 0 \), \( 3^{x} > 0 \), and \( \ln 3 > 0 \), we have \( f'(x) > 0 \) for all \( x \). Therefore the function is strictly increasing on its entire domain.

  7. Concavity.

    Compute the second derivative:
    \[ f”(x) = 2 \cdot 3^{x} \cdot (\ln 3)^{2}. \]
    This is positive for all \( x \), so the graph is concave upward everywhere.

  8. Key points to plot.

    Compute a small table of values to guide the sketch.

    x f(x) = 2 · 3^x
    -2 \( f(-2) = 2 \cdot 3^{-2} = 2 \cdot \tfrac{1}{9} = \tfrac{2}{9} \approx 0.222\)
    -1 \( f(-1) = 2 \cdot 3^{-1} = 2 \cdot \tfrac{1}{3} = \tfrac{2}{3} \approx 0.667\)
    0 \( f(0) = 2 \)
    1 \( f(1) = 2 \cdot 3 = 6 \)
    2 \( f(2) = 2 \cdot 9 = 18 \)

    Plot these points and draw a smooth curve through them that is increasing and concave up, approaching the line \( y = 0 \) on the left.

  9. Transformation description.

    Start with the basic graph of \( y = 3^{x} \), which has y-intercept 1 and horizontal asymptote \( y = 0 \). Multiplying by 2 produces a vertical stretch by factor 2. This doubles all y-values, so the y-intercept moves from \( (0,1) \) to \( (0,2) \), and other y-values are doubled accordingly.

  10. Final sketch description.

    The graph of \( f(x) = 2 \cdot 3^{x} \) is an increasing exponential curve that:

    • has domain \( (-\infty,\infty) \),
    • has range \( (0,\infty) \),
    • has horizontal asymptote \( y = 0 \),
    • passes through \( (0,2) \), \( (1,6) \), \( (-1,\tfrac{2}{3}) \),
    • is strictly increasing and concave up.

    When sketched, the curve lies very close to the x-axis for large negative x, crosses the y-axis at 2, and then rises rapidly for positive x.

See full solution

Graph

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FAQs

What type of function is f(x) = 2 \cdot 3^x?

Exponential: base 3 (>1) so the function grows as x increases; initial value (y-intercept) is 2.

What is the domain and range of f(x) = 2 \cdot 3^x?

Domain: all real numbers. Range: (0, ∞) because 2 \cdot 3^x is always positive.

What is the horizontal asymptote?

The horizontal asymptote is y = 0.

What are the x- and y-intercepts?

y-intercept: f(0) = 2 at (0, 2). There is no x-intercept because the function never equals 0.

Is the function exponential growth or decay?

Exponential growth, since the base 3 > 1; values increase multiplicatively by factor 3 for each +1 in x.

How do I sketch the graph quickly?

Plot (0,2), (1,6), (−1, 2/3), draw the horizontal asymptote y = 0, then sketch a smooth curve rising to the right and approaching 0 to the left.

What is the inverse function?

The inverse is f^{-1}(x) = \log_{3}\!\bigl(x/2\bigr), with domain x > 0.

How does f(x) relate to g(x) = 3^x (what transformation)?

f(x) = 2 \cdot 3^x is a vertical stretch of g(x) = 3^x by factor 2; asymptote y = 0 stays the same.

How do I evaluate f(x) for non-integer x?

Use exponent rules or a calculator: f(x) = 2 \cdot 3^x = 2 \cdot e^{x \ln 3}.

Where is f(x) increasing or decreasing?

f(x) is increasing for all real x because f'(x) = 2 \ln(3) \cdot 3^x > 0 for every x.
Graph f(x)=2(3)^x and analyze it now.
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