Proving the Existence of a Single Real Root Using Derivatives

[REVIANNA]

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?

 

[mfb]

The derivative is not correct, and the original function is not a cubic function due to the cosine.

how is the sin(x) value +1
does it not oscillate b/w -1 and 1?

It does oscillate, but a smaller sine value makes the inequality even stronger.
That's why the two parts are connected via $\leq$ and not =.

 

[REVIANNA]

cubic function due to the cosine.

yeah the derivative is wrong x^2 has a coefficient -18
and the inequality is correct (I was not paying attention)
does cosine(and trig fns) also change the degree? I thought that only the highest power of x indicated the degree

 

[mfb]

Cosine and sine are not powers of their argument, the function is not a polynomial function of any degree.

 

[REVIANNA]

the function is not a polynomial function of any degree.

So,How do you think about the max no of roots for the function?

 

[mfb]

It has at most one root, as shown in the first post. The proof via the derivative works for all functions with a derivative, they don't have to be polynomial functions.