How to solve $\\e^{-u}+\frac{u}{5}=1\\$ for $\\u\\$ ? What is the method to solve it without using graph. Solving $\\e^{-u}+\frac{u}{5}=1\\$ without using graph.
From graphing line $y(u)=1$ intersects with $\\e^{-u}+\frac{u}{5}\\$ at two points (0,1) &  (4.97,1) so that gives $\\u = 0\\$ or $\\4.97\\$. But how can I solve it analytically ?
[[graph]

I have come up with an "algebrical" approximate solution of the curve shown in function $\\y(u)=e^{-u}+\frac{u}{5}-1=0\\$ which is as follows:
As can be noticed and observed from graph, the curve in excess of $\\u>7.7\\$ becomes a straight line.

Hence first we take the derivative of the function $\\y(u)\\$:
$\\y(u)=e^{-u}+\frac{u}{5}-1\\\tag{1}$
$\frac{\mathrm{d(y(u)} )}{\mathrm{d} u}=-e^{-u}+\frac{1}{5}$ = slope
Now the equation of straight line as mentioned above will be:
$\\y=mu+c\\$ and $\\y'(u)\\$ becomes,
$\\y=(-e^{-u}+\frac{1}{5})u+c\\\tag{2}$
Now original equation Eq. (1) in this form will be,
$\\y=(u.e^{-u}+\frac{u}{5u})u-1\\\tag{3}$
Considering Eq. (2) & (3), it is can deduced that,
$c=(-1)$
Now put $c=(-1)$ and $y=0$ for $u-intercept$ in Eq. (2), we get:
$\\0=(e^{-u}+\frac{1}{5})u-1\\$
$\\1=u.e^{-u}+\frac{u}{5}\\$
$\\e^{-u}=\frac{1}{u}(1-\frac{u}{5})=\frac{(5-u)}{5u}\\$
Now put value of $e^{-u}$ and $y=0$ in Eq. (2) again, we get:
$\\0=(-\frac{(5-u)}{5u}+\frac{1}{5})u-1\\$ simplifying will get,
$\\5=-(5-u)+u\\$
$\\10=2u\\$
$\\u=\frac{10}{2}=5\\$
Hence $u=5$ an approximate solution.
Putting value of $u$ in Eq. (1):
$\\e^{-5}+\frac{5}{5}-1=0\\$
$\\e^{-5} = 0\\$
$\\(0.00673) \approx 0\\$
[The idea of doing all excercise is to see if there is an algebric solution available or not]
