Show that minimum value of function f(x)=e^x-x is 1.

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You need to prove that the minimum value of the function is `f(x_0) = ` `1` , hence, you need to prove that the function decreases over `(-oo,x_0)` and it increases over `(x_0,oo)` .

You need to evaluate the critical value `x_0` , hence, you need to solve for `x_0` the equation `f'(x_0) = 0` , such that:

`(e^(x_0) - x_0)' = 0 => e^(x_0) - 1 = 0 => e^(x_0) = 1 =>` `e^(x_0) = e^0 => x_0 = 0`

Replacing 0 for `x_0 ` in equation of the function yields:

`f(0) = e^0 - 0 => f(0) = 1`

Hence, evaluating the extreme value of the function yields` f(0) = 1` .

You need to test if the function decreases over `(-oo,0)` and it increases over `(0,oo)` , hence, you need to select one value `x_1` in interval `(-oo,0)` and other value `x_2 in (0,oo)` , such that:

`x_1 = -1 => f(-1) = e^(-1) - 1 = 1/e - 1 < 0`

Since the values of the function are negative over `(-oo,0)` , yields that the function decreases over` (-oo,0)` .

`x_2 = 1 => f(1) = e - 1 > 0`

Since the values of the function are positive over `(0,oo)` , yields that the function increases over `(0,oo)` .

**Hence, testing if the given function has a global minimum 1, yields that the function decreases over `(-oo,0)` and it increases over `(0,oo)` , thus, the function reaches its minimum at `x = 0` , such that **`f(0) = 1.`

We have given

`f(x)=e^x-x`

We wish to obtained minimum of f(x).

`f'(x)=e^x-1`

`Let x=c` be the cricitical point. Then

`f'(c)=0`

`e^c-1=0`

`e^c=1`

`e^c=e^0`

`c=0`

`f''(x)=e^x`

`f''(x)}_{x=c}=e^0=1 >0`

**Thus x=0 is point of minima and minimum value of f(x)=1**

Now see how its graph looks like.

See point wher f(x) and y=1 ,meet . Thus x=0 is point of minima and minimum value is 1.

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