Applying the IVT to show P(c) = 150 has a solution

First note that $P(t) = 100e^{-t/5} + t^2$ is a sum of an exponential function and a polynomial, both of which are continuous everywhere. Since a sum of continuous functions is continuous, $P$ is continuous on the closed interval $[0,6]$, so the Intermediate Value Theorem applies here.

Next evaluate the endpoints: $P(0) = 100e^{0} + 0^2 = 100(1) + 0 = 100$.

And $P(6) = 100e^{-6/5} + 6^2 = 100e^{-1.2} + 36$. Since $e^{-1.2} \approx 0.3012$, we have $P(6) \approx 100(0.3012) + 36 = 30.12 + 36 = 66.12$.

For the IVT to guarantee a solution to $P(c) = 150$, we need $150$ to lie strictly between $P(0)$ and $P(6)$. However, we have $P(0) = 100 < 150$ and $P(6) \approx 66.12 < 150$. Since $150$ is not between these two values (both are less than $150$), the Intermediate Value Theorem does not guarantee that a solution exists in the interval $(0,6)$.