Determine the x coordinate of the point for which the tangent is horizontal to the function f(x) = (2^x)/x
using the quotient rule,
f'(x) = 2^x (x ln2 - 1)/x^2
so f'(x) = 0 at x = 1/ln2
at that point,
f(1/log2) = e ln2
(because 1/ln2 = log2e)
Why did the tangent go to the function's party? Because it heard there was a lot of ex-citement! Let's find the x-coordinate where the tangent is horizontal to the function f(x) = (2^x)/x.
To find this point, we need to find where the derivative of the function equals zero since a horizontal tangent occurs when the derivative is zero.
To find the derivative of f(x), we can use the quotient rule:
f'(x) = ((x * (ln(2)*2^x)) - (2^x))/x^2
To find where the derivative equals zero, we set f'(x) = 0 and solve for x:
((x * (ln(2)*2^x)) - (2^x))/x^2 = 0
Since the numerator of the equation is zero, we have:
(x * (ln(2)*2^x)) - (2^x) = 0
We can divide both sides of the equation by 2^x, giving:
x * ln(2) - 1 = 0
x * ln(2) = 1
x = 1/ln(2)
So, the x-coordinate of the point where the tangent to the function f(x) = (2^x)/x is horizontal is approximately x ≈ 1/ln(2).
To find the x-coordinate of the point where the tangent to the function f(x) = (2^x)/x is horizontal, we need to find the critical points of the function.
Step 1: Find the derivative of the function f(x).
f'(x) = [(2^x * ln(2) * x - 2^x) / x^2]
Step 2: Set the derivative equal to zero to find the critical points.
[(2^x * ln(2) * x - 2^x) / x^2] = 0
Step 3: Simplify the equation and solve for x.
2^x * ln(2) * x - 2^x = 0
2^x * (ln(2) * x - 1) = 0
The equation 2^x = 0 has no solutions since 2 raised to any power will always be positive. Therefore, we set the expression inside the parenthesis equal to zero.
ln(2) * x - 1 = 0
ln(2) * x = 1
x = 1 / ln(2)
So, the x-coordinate of the point where the tangent is horizontal to the function f(x) = (2^x)/x is x = 1 / ln(2).
To determine the x-coordinate(s) of the point(s) where the tangent line to the function f(x) = (2^x)/x is horizontal, we need to find the critical points. A horizontal tangent line occurs when the derivative of the function is zero.
Let's start by finding the derivative of f(x). The function f(x) = (2^x)/x can be rewritten using logarithmic properties as f(x) = (e^(x * ln(2))) / x:
f(x) = e^(x * ln(2)) * x^(-1)
Using the product rule and the chain rule, we can find the derivative of f(x). Let's call the derivative g(x):
g(x) = (e^(x * ln(2)) * d/dx(x^(-1))) + (x^(-1) * d/dx(e^(x * ln(2))))
The derivative of x^(-1) is -x^(-2) and the derivative of e^(x * ln(2)) is e^(x * ln(2)) * ln(2). Substituting these values into the equation, we have:
g(x) = (e^(x * ln(2)) * (-x^(-2))) + (x^(-1) * e^(x * ln(2)) * ln(2))
Simplifying this expression, we get:
g(x) = (e^(x * ln(2)) * (-1/x^2)) + (e^(x * ln(2)) * ln(2)/x)
Next, we set g(x) equal to zero to find the critical points:
0 = (e^(x * ln(2)) * (-1/x^2)) + (e^(x * ln(2)) * ln(2)/x)
Since e^(x * ln(2)) is always positive, we can divide the equation by e^(x * ln(2)) to simplify further:
0 = -1/x^2 + ln(2)/x
Multiplying through by x^2, we have:
0 = -1 + ln(2)x
Rearranging the equation, we find:
ln(2)x = 1
Finally, solving for x, we get:
x = 1/ln(2)
Therefore, the x-coordinate of the point where the tangent line to the function f(x) = (2^x)/x is horizontal is x = 1/ln(2).