One litre of water at 0 degrees Celsius will have a different volume at higher temperatures. For a temperature 0 ≤ T ≤ 30 (in Celsius) this volume (in litres) is well approximated by the function

Question

V = −aT 3 + bT 2 − cT + 1
using the coefficients a = 9.8 × 10−8, b = 3.5 × 10−6, and c = 2.4 × 10−5.
(a) find the derivative
(b) When is the derivative positive, negative, and zero? (give answers correct to 3 decimal places)

ok, so i have a question for part b.
when u find the critical points for the derivative, do they have to be in between the interval of 0<T<30 ? because one of the critical points that i solved is not in that interval. Thank you

Answer 1

Are there supposed to be exponents in your equation??
I will assume:
V = −aT^3 + bT^2 − cT + 1 , using the given:
dV/dT = -3a T^2 + 2b T - c

I assume you plugged in the given values for
a, b, and c, used the quadratic equation formula and solved for T
Since the coefficient of T^2 is negative, you have a parabola opening downwards
I assume there are two values of T, which are the x-intercepts.
Yes, you should only use the values of T that fall in the given domain of 0≤T≤30
Find the T value of the vertex.
The derivative will be negative for values less than that T.
It will be zero for the T of the vertex
and it will be positive for values > T of the vertex

I will not do the arithmetic for you, my calculator is not handy right now.