Can this be contradicted? (1 Viewer)

[leehuan]

If f' is (continuous, differentiable and) monotone, f always concaves the same way

 

[InteGrand]

If f' is (continuous, differentiable and) monotone, f always concaves the same way

If f' is differentiable and monotone on an interval I, we must have (f')' = f" stay the same sign (possible 0 some places) over that interval. Let's just suppose it's always non-negative.

Since f is twice differentiable, the sign of f" dictates the concavity of f over the interval. Since f" is always non-negative on I, f is convex on the interval I.

If these properties of f and its derivatives are true on R, then f will be convex on R.

We'd change convex to concave if f" were non-positive instead.

 

[leehuan]

Thanks but cause I just realised that questioning that wasn't enough, I'm just going to send the question



Previous parts:

i) Show that p3(x)=1+x+x^2/2!+x^3/3! has at least one real root. Done by intermediate value theorem
ii) Show that p2(x)=1+x+x^2/2! has no real roots and deduce p3(x) has exactly one real root.

Although p2(x) could have no real roots can be easily done by the quadratic discriminant and finding that it's positive definite, using the method taught in class I did:
p2'(x)=1+x has one zero at x=-1
So p2(x) has at most 1 root
Hence, Test the signs of p2 for the intervals (-inf, -1] and [1, inf) as between any two stationary points (or one stationary point and inf.) there can only be at most one root. (I think this was a corollary of Rolle's theorem.)
On both intervals, p2 was positive so p2 > 0 for all x

So because p2 has no roots, noting that p3'(x)=p2(x), p3(x) has at most one zero.

So p3 has exactly one zero.

 

[InteGrand]

Thanks but cause I just realised that questioning that wasn't enough, I'm just going to send the question



Previous parts:

i) Show that p3(x)=1+x+x^2/2!+x^3/3! has at least one real root. Done by intermediate value theorem
ii) Show that p2(x)=1+x+x^2/2! has no real roots and deduce p3(x) has exactly one real root.

Although p2(x) could have no real roots can be easily done by the quadratic discriminant and finding that it's positive definite, using the method taught in class I did:
p2'(x)=1+x has one zero at x=-1
So p2(x) has at most 1 root
Hence, Test the signs of p2 for the intervals (-inf, -1] and [1, inf) as between any two stationary points (or one stationary point and inf.) there can only be at most one root. (I think this was a corollary of Rolle's theorem.)
On both intervals, p2 was positive so p2 > 0 for all x

So because p2 has no roots, noting that p3'(x)=p2(x), p3(x) has at most one zero.

So p3 has exactly one zero.

Note that p₄'(x) = p₃(x) for all x. Also, p₄"(x) = p₂(x) for all x. We've shown that p₂(x) is a positive definite quadratic, so that means the second derivative of p₄ is always positive. This means the minimum of p₄ will be a global min.

Now, to find this min., set p₄'(x) = p₃(x) = 0. (There is exactly one solution to this, as you've shown in a previous part.)

But observe that p₄(x) = p₃(x) + x⁴/4!. So at the point where p₄'(x) ≡ p₃(x) = 0, say at x = a, we have p₄(a) = 0 + a⁴/4! > 0 (strictly positive since clearly a is not zero, because p₃ doesn't have its root at 0).

So the absolute minimum value of p₄(x) is positive. Hence p₄(x) > 0 for all real x.