Find $f$ such that $f^{\prime}(x)=\frac{8}{\sqrt{x}}, f(16)=78$
\[
f(x)=
\]

Step 1 ：The problem is asking for a function \(f(x)\) such that its derivative \(f'(x)\) is equal to \(\frac{8}{\sqrt{x}}\) and \(f(16)\) is equal to 78. This is a problem of finding the antiderivative (or integral) of a function and then using the given point to solve for the constant of integration.

Step 2 ：The antiderivative of \(\frac{8}{\sqrt{x}}\) can be found using the power rule for integration, which states that the integral of \(x^n\) is \(\frac{1}{n+1}x^{n+1}\), where \(n\) is any real number except -1. In this case, we can rewrite \(\frac{8}{\sqrt{x}}\) as \(8x^{-1/2}\), so the antiderivative will be \(\frac{1}{-1/2+1}8x^{-1/2+1}\).

Step 3 ：After finding the antiderivative, we can use the given point \((16, 78)\) to solve for the constant of integration. This is done by substituting the x and y values into the antiderivative equation and solving for the constant.

Step 4 ：The function \(f(x)\) that satisfies the given conditions is \(f(x) = 16\sqrt{x} + 14\).

Step 5 ：Final Answer: \(\boxed{f(x) = 16\sqrt{x} + 14}\)