Compute for each of the following continuous functions .

- .
- .
- .
- .

- Since
we take the derivative of both sides to obtain

Therefore,

- Since
we take the derivative of both sides (see part (b) of this exercise for taking the derivative of an integral like this) to obtain

Thus,

- Here we have
Here we evaluate the integral on the left,

Therefore,

Thus,

- Finally, since
we take the derivative of both sides,

Then, since

(since the cubic polynomial has only one real root at ) we have

Technically x can be zero…

Is it necessary to point out that the polynomial x^3 + x^2 = 2 has only one (real) solution? It seems to me that we have already shown that f(x^3+x^2)=1/(3x^2+2x) for all x >= 0. The number 1 is then simply one such number that just happens to give us the particular solution that we are looking for. If there was another y != x such that y^3 + y^2 = 2, it seems to me like it shouldn’t invalidate our solution.