Lösung 1.3:3b
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Version vom 10:11, 11. Mär. 2009
Because the function is defined and differentiable for all x, the function can only have local extreme points at the critical points, i.e. where the derivative is zero.
For this function, the derivative is given by
| \displaystyle f^{\,\prime}(x) = -3e^{-3x} + 5 |
and if we set it to zero, we will obtain
| \displaystyle 3e^{-3x} = 5 |
which is a first-degree equation in \displaystyle e^{-3x} and has the solution
| \displaystyle x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.} |
The function therefore has a critical point \displaystyle x=-\frac{1}{3}\ln \frac{5}{3}\,\textrm{.}
Instead of studying the sign changes of the derivative around the critical point in order to decide if the point is a local maximum, minimum or neither, we investigate the function's second derivative.
The second derivative is equal to
| \displaystyle f^{\,\prime\prime}(x) = -3\cdot (-3)e^{-3x} = 9e^{-3x} |
and is positive for all values of x, since the exponential function is always positive.
In particular, this means that
| \displaystyle f^{\,\prime\prime}\Bigl( -\frac{1}{3}\ln \frac{5}{3} \Bigr) > 0\,, |
which means that \displaystyle x=-\tfrac{1}{3}\ln\tfrac{5}{3} is a local minimum.
