To find the maximum value of the function \(f(t) = e^{-t} - 2e^{-2t}\) for \(0 \le t < \infty\), we use differential calculus. The process involves finding the first derivative of the function, setting it to zero to find critical points, and then using the second derivative test to confirm if these points correspond to a maximum.

Function Analysis: Initial Steps

The given function for which we need to find the maximum value is:

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

The domain for \(t\) is specified as \(0 \le t < \infty\).

Derivative Calculation

To find the maximum or minimum values of a function, a fundamental step is to calculate its first derivative, \(f'(t)\), and set it equal to zero. This helps us locate the critical points.

Let's differentiate \(f(t)\) with respect to \(t\):

\(f'(t) = \frac{d}{dt}(e^{-t} - 2e^{-2t})\)

Using the chain rule, which states that the derivative of \(e^{ax}\) is \(ae^{ax}\):

• The derivative of \(e^{-t}\) is \((-1)e^{-t} = -e^{-t}\).
• The derivative of \(2e^{-2t}\) is \(2(-2)e^{-2t} = -4e^{-2t}\).

Combining these, the first derivative \(f'(t)\) is:

\(f'(t) = -e^{-t} - (-4e^{-2t})\)

\(f'(t) = -e^{-t} + 4e^{-2t}\)

Critical Point Determination

To determine the critical points where the function might attain its maximum value, we set the first derivative \(f'(t)\) equal to zero:

\(f'(t) = 0\)

\(-e^{-t} + 4e^{-2t} = 0\)

Now, let's rearrange this equation to solve for \(t\):

\(4e^{-2t} = e^{-t}\)

Since \(e^{-2t}\) is always positive (never zero), we can safely divide both sides of the equation by \(e^{-2t}\):

\(\frac{4e^{-2t}}{e^{-2t}} = \frac{e^{-t}}{e^{-2t}}\)

\(4 = e^{-t - (-2t)}\)

\(4 = e^{-t + 2t}\)

\(4 = e^{t}\)

To isolate \(t\), we take the natural logarithm (\(\log_e\) or \(\ln\)) of both sides of the equation:

\(\log_e 4 = \log_e (e^t)\)

Using the property \(\log_b (b^x) = x\):

\(\log_e 4 = t\)

So, the critical point for the function is \(t = \log_e 4\).

Maximum Value Confirmation

To confirm that this critical point \(t = \log_e 4\) corresponds to a maximum value, we can use the second derivative test. First, we need to calculate the second derivative, \(f''(t)\).

\(f''(t) = \frac{d}{dt}(-e^{-t} + 4e^{-2t})\)

Differentiating each term from the first derivative \(f'(t)\):

• The derivative of \(-e^{-t}\) is \(-(-e^{-t}) = e^{-t}\).
• The derivative of \(4e^{-2t}\) is \(4(-2)e^{-2t} = -8e^{-2t}\).

Thus, the second derivative \(f''(t)\) is:

\(f''(t) = e^{-t} - 8e^{-2t}\)

Now, we substitute the critical point \(t = \log_e 4\) into \(f''(t)\):

\(f''(\log_e 4) = e^{-(\log_e 4)} - 8e^{-2(\log_e 4)}\)

Using the logarithm property \(e^{\ln x} = x\) and \(e^{-n \ln x} = e^{\ln(x^{-n})} = x^{-n}\):

\(f''(\log_e 4) = \frac{1}{4} - 8e^{\log_e (4^{-2})}\)

\(f''(\log_e 4) = \frac{1}{4} - 8e^{\log_e (1/16)}\)

\(f''(\log_e 4) = \frac{1}{4} - 8\left(\frac{1}{16}\right)\)

\(f''(\log_e 4) = \frac{1}{4} - \frac{8}{16}\)

\(f''(\log_e 4) = \frac{1}{4} - \frac{1}{2}\)

\(f''(\log_e 4) = \frac{2-4}{8} = -\frac{2}{8} = -\frac{1}{4}\)

Since \(f''(\log_e 4) = -\frac{1}{4} < 0\), according to the second derivative test, the function \(f(t)\) has a local maximum at \(t = \log_e 4\).

Boundary Condition and Limiting Behavior

To ensure this is the absolute maximum, we also need to consider the function's value at the boundary of the domain and its behavior as \(t\) approaches infinity.

• At the boundary \(t=0\):

  \(f(0) = e^{-0} - 2e^{-2(0)}\)

  \(f(0) = e^0 - 2e^0\)

  \(f(0) = 1 - 2(1)\)

  \(f(0) = 1 - 2 = -1\)

• As \(t \to \infty\):

  \(\lim_{t \to \infty} f(t) = \lim_{t \to \infty} (e^{-t} - 2e^{-2t})\)

  As \(t\) approaches infinity, both \(e^{-t}\) and \(e^{-2t}\) approach 0.

  \(\lim_{t \to \infty} f(t) = 0 - 2(0) = 0\)

Comparison of Values

Let's summarize the values of the function at the critical point and boundary points:

Comparing the values, \(-1\), \(\frac{1}{8}\), and \(0\), the largest value is \(\frac{1}{8}\), which occurs at \(t = \log_e 4\). This confirms that the global maximum for \(f(t)\) in the given domain occurs at this point.

Therefore, the maximum value of the function \(f(t) = e^{-t} - 2e^{-2t}\) occurs at \(t = \log_e 4\).