Question

Curvature of \(e^{x}\) Find the curvature of \(f(x)=e^{x}\) and find the point at which it is a maximum. What is the value of the maximum curvature?

Short answer

Answer: The curvature of the function \(f(x) = e^x\) does not have a maximum value.

Step-by-step solution

Find the first and second derivatives of f(x) = e^x

To find the curvature, we first need to find the first and second derivatives of the given function. In our case, f(x) = e^x. The first derivative, f'(x), is the derivative of e^x with respect to x which is e^x itself. Now, we find the second derivative, f''(x), which is the derivative of the first derivative with respect to x. Since the first derivative is e^x, the second derivative is also e^x. So, f'(x) = e^x and f''(x) = e^x.

Find the curvature using the curvature formula

The curvature, K, of a function at a certain point can be found using the following formula: \[K(x) = \frac{|f''(x)|}{(1 + (f'(x))^2)^{3/2}}\] Substitute our first and second derivatives into this formula: \[K(x)= \frac{|e^x|}{(1 + (e^x)^2)^{3/2}}\]

Find the maximum curvature and its point

To find the maximum curvature, we need to find the critical points by taking the derivative of K(x) with respect to x and then set it equal to 0. First, let's simplify the curvature formula: \[K(x) = \frac{e^x}{(1 + e^{2x})^{3/2}}\] Now differentiate K(x) with respect to x: \[K'(x) = \frac{e^{2x}(1 + e^{2x})^{3/2} - 3e^{3x}(1 + e^{2x})^{1/2}}{(1 + e^{2x})^3}\] Set K'(x) = 0: \[\frac{e^{2x}(1 + e^{2x})^{3/2} - 3e^{3x}(1 + e^{2x})^{1/2}}{(1 + e^{2x})^3} = 0\] The denominator can be ignored, since it will never be zero. \[e^{2x}(1 + e^{2x})^{3/2} - 3e^{3x}(1 + e^{2x})^{1/2} = 0\] Finding the solutions to this equation is quite difficult, but since the function e^x is always positive, K'(x) is never zero. Therefore, the curvature has no critical points and no maximum value. In conclusion, the curvature of the function \(f(x) = e^{x}\) does not have a maximum value.