In my Physics Textbook, I've been asked to denote the maximum or minimum value of this function (whichever is applicable):
$$x(t) = x_0(1 - e^{-\gamma t}), \text{where } t \ge 0, x_0 > 0$$
Now, I tried to do it like this.
First, I tried to find a point in the graph of the function which has the slope $0$. My logic is the maxima or minima, both of the points have slope $0$, so by trying to equate the derivative w.r.t $t$ with $0$, we can have the corresponding value for t.
But, I failed, horribly so.
$$\frac{dx}{dt} = \frac{d}{dt}[x_0(1 - e^{-\gamma t})]$$
$$= x_0\frac{d}{dt}(1 - e^{-\gamma t})$$
$$= x_0[-e^{-\gamma t} * \frac{d}{dt}(-\gamma t)]$$
$$= x_0(-e^{-\gamma t} * -\gamma) = \gamma x_0e^{-\gamma t}$$
Trying to equate this with $0$,
$$\gamma x_0e^{-\gamma t} = 0$$
$$\implies e^{-\gamma t} = 0 \text{[As } \gamma \text{ and } x_0 \text{ both are constants]}$$
And wow, that is one hard slap to the face. Great equation to solve, hmm (Self-depreciating chuckle).
Please, help me.
Essentially what is going on here is that your function $x(t)$ is always increasing (for $\gamma>0$), so there is no finite point at which $\frac{dx}{dt}=0$. This means that the maximum and minimum values must occur at the end points of your interval, $t\in[0,\infty)$.
Since $x(t)$ is increasing, its minimum occurs at $t=0$ and is given by $x(0)=0$. On the other hand, since your interval is infinite, $x(t)$ never actually attains a maximum value. We do however have that $\lim_{t\rightarrow\infty}x(t)=1$, that is $x(t)$ approaches $1$ (but never reaches it in finite time).