# Determine the x coordinate of the point for which the tangent is horizontal to the function f(x)= 2^x/x.

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Determine the x-coordinate of the point for which the tangent is horizontal to the function `f(x)=(2^x)/x` :

The tangent to a function will be horizontal if the first derivative is zero at the point of tangency. So we need only determine the critical points of the function. (The critical points of a function are the points where the first derivative is zero or fails to exist -- these are the only places the function can have a minimum or maximum, which are the only places for a horizontal tangent line)

`f'(x)=(x(ln2)2^x-2^x)/(x^2)` We use the quotient rule:

`d/(dx)((f(x))/(g(x)))=(g(x)f'(x)-f(x)g'(x))/((g(x))^2)` and `d/(dx)a^x=(lna)a^x`

We can simplify the expression a little:

`f'(x)=(2^x(xln2-1))/x^2`

Now for `f'(x)=0` the numerator must be zero, while the denominator is nonzero. We note that for x=0 the numerator is -1 so we need only check for the numerator to be zero:

`2^x(xln2-1)=0` By the zero product property either `2^x=0` , which cannot happen, or `xln2-1=0`

`=>xln2=1`

`=>x=1/(ln2)~~1.443`

**Thus the x-value where the tangent to f(x) is horizontal is** `x=1/(ln2)`

This corresponds to a local minimum:

The tangent is a horizontal, which means that it is of the form y=b, hence the slope of that line is zero.

So we need to find that first derivative and find the value of x that makes it zero.

`f'(x)=[2^x*ln2*x-2^x*1]/x^2 =>`

`f'(x)=[2^x(xln2-1)]/x^2 `

` ``f'(x)=0 => xln2-1=0 =>`

`x=1/ln2`

**Thus the x coordinate for that point is 1/ln2**