For 0 ≤ 𝑡 < ∞, the maximum value of the function (𝑡) = 𝑒 −𝑡 −

SOLUTION

Concept:

The point of maxima or minima is obtained by solving for the derivative of the function and equating to zero.

Then, to check if the point is a point of maxima ‘or’ minima we check the second derivative at that point.

This is explained with the help of the following graph:

If \(\frac{{{d^2}f}}{{d{x^2}}} < 0\); the point will be a point of maxima

If \(\frac{{{d^2}f}}{{d{x^2}}} > 0\), the point will be a point of minima.

Analysis:

Given f(t) = e^-t – 2e^-2t

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

Solving for f’(t) = 0, we get:

4e^-2t = e^-t

\(\frac{4}{{{e^t} \cdot {e^t}}} = \frac{1}{{{e^t}}}\) 

e^t = 4

t = In 4

f”(t) = e^-t – 8 e^-2t

At t = In 4, we get

f”(In 4) = e^{-In 4} – 8 e^{-2 In 4}

\( = \frac{1}{4} - \frac{8}{{16}}\) 

\( = \frac{1}{4} - \frac{1}{2}\) 

= -0.26

Since, f”(t) < 0 at t = In 4, it is a point of maxima.