Find an equation for the tangent to the curve at the given point.
7) $f(x)=10 \sqrt{x}-x+7,(100,7)$

Step 1 ：Given the function \(f(x) = 10 \sqrt{x} - x + 7\) and the point \((100,7)\).

Step 2 ：First, we need to find the derivative of the function \(f(x)\).

Step 3 ：The derivative of \(f(x)\) is \(f'(x) = -1 + \frac{5}{\sqrt{x}}\).

Step 4 ：Substitute the x-coordinate of the given point into the derivative to find the slope of the tangent line. So, \(m = f'(100) = -1 + \frac{5}{\sqrt{100}} = -\frac{1}{2}\).

Step 5 ：Then, we use the point-slope form of the line equation to find the equation of the tangent line. The point-slope form is \(y - y_1 = m(x - x_1)\).

Step 6 ：Substitute the given point \((100,7)\) and the slope \(m = -\frac{1}{2}\) into the point-slope form, we get \(y - 7 = -\frac{1}{2}(x - 100)\).

Step 7 ：Solve for y to get the equation of the tangent line: \(y = 57 - \frac{x}{2}\).

Step 8 ：Final Answer: The equation of the tangent line to the curve \(f(x)=10 \sqrt{x}-x+7\) at the point \((100,7)\) is \(\boxed{y = 57 - \frac{x}{2}}\).