A 3rd degree function "f" satisfies the following:
f'(2)=-1
f''(4)=0

Calculate f'(6).

Question

A 3rd degree function "f" satisfies the following:
f'(2)=-1
f''(4)=0

Calculate f'(6).

imqwerty · Answer

f' is the derivative of f right?

owlcoffee · Answer

yeah, I believe so.
I found this and I am skeptic if the given information is sufficient to deduce that...

anonymous · Answer

The key is the phrase "third degree function", which I assume means "third degree polynomial".  If that's the case, then the information is sufficient.

owlcoffee · Answer

Well, the derivative of a 3rd degree polynomial becomes a 2nd degree polynomial, which would imply that two, one or none extrema should exist.

anonymous · Answer

Because there is never any reference to f itself, we may as well consider a second degree function g = f' such that
g(2) = -1
g'(4) = 0

That implies that g has either a minimum or a maximum at x=4.  In either case, g(6) should be equal to g(2), so the answer to the question is

g(6) = f'(6) = -1.

imqwerty · Answer

ok its a 3rd degree function
so
lets say$$f(x)=ax^3+bx^2+cx+d$$

now$$f'(x)=3ax^2+2bx+c$$$$f'(2)=3a(2^2)+2b(2)+x=-1 $$

so u have 1st equation as->
$$12a+4b+c=-1$$

now$$f''(x)=6ax+2b$$$$f''(4)=6a(4)+2b=0$$$24a+2b=0$    -->2nd equation

now we need to find f''(6)
$$f'(6)=108a+12b+c$$

now just try to add eq 1  with 4 times of equation 2 :D

owlcoffee · Answer

yeah, I found that as a possible process of solution, I'm guessing that finding the function is the way to go, I thought that there could be any generalizations possible from this (I haven't touched derivatives in a while now).

imqwerty · Answer

$\Huge\ddot\smile$

owlcoffee · Answer

By the way, I think it becomes a problem, since it'll be two equations with three variables.

imqwerty · Answer

We need to find ---->$f'(6)=108a+12b+c$

1st equation-------->$-1=12a+4b+c$

2nd equation------->$0=24a+2b$

adding the eq 1 to 4 times of equation 2 u get this->

$-1+4(0)=12a+4b+c+4(24a+2b)$
$-1=108a+12b+c$
but
$f'(6)=108a+12b+c$

so$f'(6)=108a+12b+c=-1$

owlcoffee · Answer

Ah, I see, you worked the algebra to make their expressions fit, and thus one being equal to the other.

imqwerty · Answer

e.e yes!

parthkohli · Answer

If we let$$f''(x) = k(x - 4)$$and step backwards, then we get $f'(x) = kx^2/2 - 4kx + c$. Now $f'(2) =-1$ so $c =  6k-1$. In this case, we get $f'(x) = kx^2/2 - 4kx + 6k-1$ and so $f'(6) = -1$.