The point P (4, –1) lies on the curve C with equation y = f( x ), x > 0, and f '(x) =x/2 - 6/√x + 3. Find the equation of the tangent to C at the point P , giving your answer in the form y = mx + c. Find f(x)

First step is to understand all the notation, i.e. the prime for derivative. We then need to understand that the derivative is the gradient of the tangent at a point. In this case at the point P we can find the gradient, m, for the tangent by simply plugging in the x value to f', giving us m=2. This gives us y = 2x + c. To find c we just need a point that lies on the tangent, in this case P, and we substitue in and rearrange for c, giving c=-9. Resulting in the final equation, y=2x-9. The second part is simple integration. We need to put f' into exponent form, f'=x/2 - 6x^(-1/2) + 3. Now using our method of integration we get, f = 1/4 x^2 - 12x^(1/2) + 3x + c, remembering the constant of integration c. Final step is to find the constant of integration. Substituting in point P we get, -1 = 4 - 24 + 12 + c, which gives c = 7.