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:
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`
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)=0 => xln2-1=0 =>`
Thus the x coordinate for that point is 1/ln2