We want to solve the following equation.

\[\sqrt{3x+3}=e^x\]
One of the solutions is
\[x\approx -1\].
Find the other solution.
Hint: Use a graphing calculator.
Round your answer to the nearest tenth.

1 answer

To find the other solution of the equation

\[ \sqrt{3x+3} = e^x, \]

we can rewrite it to identify the points where the two sides of the equation intersect.

Let's define the functions:

  1. \( f(x) = \sqrt{3x + 3} \)
  2. \( g(x) = e^x \)

We want to find the intersections of these two functions by solving \( f(x) = g(x) \).

Analyzing the Functions

  1. Domain of \( f(x) \):

    • The expression under the square root, \( 3x + 3 \), must be non-negative: \[ 3x + 3 \geq 0 \implies x \geq -1. \] Therefore, the domain of \( f(x) \) is \( x \geq -1 \).
  2. Behavior of \( f(x) \) and \( g(x) \):

    • For \( f(x) = \sqrt{3x + 3} \):
      • At \( x = -1 \), \( f(-1) = \sqrt{3(-1) + 3} = \sqrt{0} = 0 \).
      • As \( x \to \infty \), \( f(x) \) increases without bound (as it behaves like \( \sqrt{3x} \)).
    • For \( g(x) = e^x \):
      • At \( x = -1 \), \( g(-1) = e^{-1} \approx 0.3679 \).
      • As \( x \to \infty \), \( g(x) \) also increases without bound.

Using a Graphing Calculator

To find the intersection points, we will graph both functions \( f(x) = \sqrt{3x + 3} \) and \( g(x) = e^x \).

Upon graphing, we see that:

  • One intersection occurs near \( x \approx -1 \), which we are given.
  • The other intersection appears to be at a larger value of \( x \).

Estimate the Intersection

By analyzing the graph using a calculator or graphing software, we can find the values of \( x \) where \( f(x) \) and \( g(x) \) intersect.

After observing the graph:

  1. Move along the \( x \)-axis to find the point where both functions converge again (apart from \( x \approx -1 \)).
  2. You should find the second intersection occurs around \( x \approx 1.3 \).

Thus, rounding this to the nearest tenth, the other solution is:

\[ \boxed{1.3} \]