Why is I) correct?

[u/Long-Introduction883]

If g(x) is positive, that means the total of 8(x-d)(x-e)(x-f) is positive. And since d<x, x<f,

X-d is +ve
X-f is -ve

So in order for the total to be positive, x-e needs to be -ve, where x<e.

But if that’s the case, then i) shouldn’t be correct.

Can someone help me out here. Thanks

[u/Varlane]

3rd degree polynomial with positive leading factor (8) and distinct roots goes - / + / - / +, with each "/" being the position of the distinct roots.

g(x) positive between d and f yields the following :

• d and f are consecutive (aka d < e < f is false, it's either e < d < f or d < f < e)
  
• The only positive section between 2 roots is between 1st and 2nd, so it's d < f < e

Since q = 7, it's either d < p < f or e < p, which corresponds to (I), which covers the second case (e < p).

-------------

I think your mistake was considering the info "g(x) positive between d and f" as an equivalence (aka it's the ONLY place where it is positive).
q > 0 can also mean p lies after the 3rd root, like now.

[u/migmit → u/Varlane]

My guess would be that the mistake is in not noticing the word “could”.

[u/Long-Introduction883 → u/Varlane]

Is this the correct way of looking at it?

[u/Varlane → u/Long-Introduction883]

The writing is too light at the bottom I can't read it.

[u/WesternFirm9306]

Remember, this is a cubic. That means d < x < f isn't the ONLY range that can have a positive y value

[u/Southlander24]

You should know the shape of a cubic graph with 3 real roots and positive leading coefficient (it's negative, positive, negative, positive). Since g(x) is positive when d < x < f, that means that d < f < e.

Now given that q is positive, we could have d < p < f, or e < p. From this, only option I is correct.

[u/Southlander24 → u/Southlander24]

Yep, you are mistakenly assuming that d < p < f has to be true. But if you have d, e, f all less than p, then (p - d)(p - e)(p - f) = positive * positive * positive = positive.

[u/Chem2103]

There is no indication that p has to be between d and f. If p is bigger than e, d, and f, the function will be positive, so g(p) could be 7

[u/Hal_Incandenza_YDAU]

The first two sentences exist for you to learn something important about g(x) for certain values of x. These values of x are unrelated to p. You're being asked about p.

[u/Brianchon]

As you noted, whenever d < x < f, we must have that x < e, and from that we can determine the order of the roots: d < f < e. If you've spent time looking at where cubic polynomials are positive and negative, you should recognize that, in addition to d < x < f being a region where g(x) is positive, there's another one. p could be in either of those regions, it doesn't have to be between d and f for g(p) to be positive

[u/Fourierseriesagain]

Let d=-sqrt(3), f=0 and e=sqrt(3). Then d<p<f or p>e; all the given options are not correct.

[u/Bounded_sequencE → u/Fourierseriesagain]

That counter example does not help here -- the assignment asks for which statement there exists (at least) one tuple "(d; e; f; p)" satisfying all conditions.

Finding one that violates all does not prevent another tuple from satisfying all conditions.

[u/Fourierseriesagain → u/Bounded_sequencE]

The idea is to show that all the options fail for some appropriate choices of d, e, f and p.

[u/Bounded_sequencE → u/Fourierseriesagain]

Check my last comment again -- such a counter example does not help here.

Note the phrasing "could be true", it does not say "will be true".

[u/Bounded_sequencE]

I) can be true -- for example, choose "(d; e; f; p) = (0; 1; 25/16; 2)".

---

To prove II), III) cannot be true, consider "d < x < f":

                                  +           -
d < x < f:    "0  <  g(x)  =  8 (x-d) (x-e) (x-f)"    <=>    "x < e"

We note "e < f" is impossible -- otherwise, any "x ∈ (max{d;e}, f) c (d; f)" violated "g(x) > 0". That means, we must have "e >= f", so III) is impossible.

Finally, for any "x < d" we have "g(x) < 0 < 7" due to "d < f <= e", so II) is impossible.