Why there is multiple root? My teacher said that 
If we have$$ f(x)=x^4 $$
Then there will be 4 same root $0$ satisfying the equation .
He said that it is because
$$f'(x)=4x^3$$
$$f''(x)=12x^2$$
$$f'''(x)=24x^1$$
 All are zero at $0$
Another example of such type of question is 
$$e^{x-1} -x = 0$$
I want to know why this happen 
 A: The equation $e^{x-1}-x=0$ has a double root at $x=1$ because:
for $x=1$ we have 
$y=e^{x-1}-x=\frac{e^x}{e}-x=\frac{e}{e}-1=0$ and also:
$$
y'=\frac{e^x}{e}-1=\frac{e}{e}-1=0
$$

We say that a function $f(x)$ has a double root at $x=a$ if it can be factorized as $f(x)=(x-a)^2h(x)$  and, in this case we have
$$
f'(x)=2(x-a)h(x)+(x-a)^2h'(x)
$$
so that also $f'(a)=0$
On the other side, you can use Taylor series to show that if $f(a)=f'(a)=0$ than $f(x)$ can be factorized as  $f(x)=(x-a)^2h(x)$.
A: Read this:
Multiplicity of a root of a polynomial
(especially the answer by Bernard)
to find some information about multiple roots.
For $e^{x-1}-x$: what are you asking about? It indeed vanishes at $x=1$ and the same happens to the derivative, but "multiple root" refers to polynomials.
A: Every polynomial of degree 4 can be decomposed as $c(x-a_1)(x-a_2)(x-a_3)(x-a_4)$, where the $a_i$ are the roots of the polynomial.
It may happen that not all $a_i$ are distinct from each other, in which case one says one has a multiple root. 
In particular this is the case for $x^4=(x-0)(x-0)(x-0)(x-0)$ which has $0$ as a 4-fold root.
A: I think:
For a polynomial, $p(x) = a_nx^n + ..... + a_1x + a_0$ we say that $b$ is a root if $p(b) = 0$.  This happens if and only if $p(x) = (x -b)(c_{n-1}x^{n-1} + .... + c_1x + c_0)$ where $(c_{n-1}x^{n-1} + .... + c_1x + c_0)$ is another polynomial.
Example $2,-2, 3$ are all roots to $p(x)= x^3 -3x^2 - 4x + 12$ and $p(x) = (x-2)(x+2)(x-3)$.
We say $b$ is a "multiple" root if $p(b) = 0$ and $p(b) = (x-b)(x-b)(c_{n-2}x^{n-2} + .... + c_1x + c_0)$ because $(x -b)$ does just divide into $p(x)$ once; it divides into it multiple times.
Now.  Here is a neat thing about multiple roots of polynomials:
If $b$ is an "k multiple root of $p(x)$ that means $p(x) = (x - b)^kq(x)$ for some other polynomial $q(x)$.  So $p'(x) = k(x - b)^{k-1}q(x) + (x - b)^kq'(x)$ and so $p'(b) = k(b-b)^{k-1}q(b) + (b -b)^kq'(b) = 0$.  So not only is $p(b) = 0$ but $p'(b) = 0$. 
We can repeat this down to the n-1 derivative.
Okay, that was because $p(x)$ is a polynomial.
$f(x) = e^{x-1} - x$ is not a polynomial.  We can not factor $f(x)$. (Not for any significant purpose at any rate.)  But $f(1) = e^{1-1} - 1 = 1-1 = 0$ so $1$ is a root of $f(x)$ because $f(1) = 0$.  
Furthermore $f'(x) = e^{x-1} - 1$ and $f'(1) = e^{1-1} -1 = 0$.  So both $f(1) = f'(1)=0$ so $1$ is a root of both $f(x)$ and $f'(x)$.  Just as a double root for a polynomial was.
Although I haven't ever heard anyone call $1$ a "double root" of $f(x)$ but maybe some instructors do refer to "double roots" meaning $f(x) = f'(x) = 0$.
A: So, if I can understand you correctly, you want to know how we can have a function $f$ and a number $a$ so that $f(a) = f'(a) = 0$. The reason is exactly what it says in the equation: In addition to the funciton value being $0$, the derivative is also $0$, which means that the graph of the function aligns itself with the $x$-axis. So the graph not only crosses the $x$ -axis, but is actually tangent to it.
