- #1
REVIANNA
- 71
- 1
Homework Statement
the original function is ##−6 x^3−3x−2 cosx##
##f′(x)=−2x^2−3+2sin(x)##
##−2x^2 ≤ 0## for all x
and ##−3+2 sin(x) ≤ −3+2 = −1##, for all x
⇒ f′(x) ≤ −1 < 0 for all x
The Attempt at a Solution
this problem is part of a larger problem which says
there is a cubic function which can have at least one real roots
than we prove (like the above) that the derivative is negative and therefore the function is strictly decreasing (so it cannot intersect the x-axis again to have another root coz it cannot increase)
therefore it has exactly one real root (not three)
my problem is how they proved that ##-3+2 sin(x)## is -1
how is the sin(x) value +1
does it not oscillate b/w -1 and 1?