if f(x)=4x^3 +24x^2 +36x +k. the function f has a local min at x=-1, and the graph of f has a point of inflection at x=-2. if integral [0, 1] f(x)dx=32, what is the value of k?? please show steps!

Question

dape · Answer

Okay, so we get some values of the derivatives (local min, inflection point), but that won't help us determine k since constants disappears when we take derivatives. The last bit, however, states that
$$ \int_0^1{f(x)dx}=32 $$
or if we put in the value of f(x),
$$ \int_0^1{(4x^3+24x^2+36x+k)dx} \
= x^4+8x^3+18x^2+kx|_0^1 \
= (1^4+8*1^3+18*1^2+k*1)-(0+0+0+0) = k+27 $$
This should be equal to 32 by the equation above, or
$$  k+27=32 $$
gives
$$ k=5 $$