A curve with equation y=f(x) passes through the point (1, 4/3). Given that f'(x) = x^3 + 2*x^0.5 + 8, find f(x).

We know that f'(x) = x^3 + 2*x^0.5 + 8, and we can integrate both sides. This gives us f(x) = (1/4)*x^4 + (4/3)x^(3/2) + 8x + c, remembering to add the constant of integration. Now, we are almost there, but we need to use the first bit of information from the question. The curve y=f(x) passes through (1, 4/3), so when x=1, y=4/3. This is the same as saying that when x=1, f(x)=4/3, i.e. f(1)=4/3. But we know that f(1)= (1/4)*1^4 + (4/3)1^(3/2) + 81 + c, from before (just put 1 instead of x), so setting this equal to 4/3 gives: 1/4 + 4/3 + 8 + c = 4/3, 1/4 + 8 + c = 0, 33/4 + c = 0, c = -33/4. So f(x) = (1/4)*x^4 + (4/3)x^(3/2) + 8x - 33/4.