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.