I Need help with concept based questions

[Deebu R]

Let f(x) and g(x) be twice differentiable functions on [0,2] satisfying f"(x)=g"(x), f'(1)=4, g'(1)=6,f(2)=3 and g(2)=9.Then what is f(x)-g(x) at x=4 equal to?

I honestly have no idea how to do it.Please help.

 

[Deebu R]

I tried to do it myself and I got -6. This is actually a prev year question so when I checked online the answer was published as 2.

 

[micromass]

The question makes no sense. You say the domain is $[0,2]$ but then ask the value at $4$?

 

[Deebu R]

The question makes no sense. You say the domain is $[0,2]$ but then ask the value at $4$?

No. The value of f(x)-g(x) if x=4.
So value of f(4)-g(4).

 

[micromass]

$f(4)$ makes no sense if the domain is $[0,2]$.

 

[member 587159]

This belongs in the homework section.

Also, you cannot evaluate the function at x = 4. However, I assume that's some kind of typo.

Start by integrating $f(x) = g(x)$ twice.
What do you know? Keep in mind: $f' = g' \Rightarrow f = g + c$

 

[Deebu R]

So, f"(x)=g"(x)
Integrating, f'(x)=g'(x)+c ===> (A)
f(x)=g(x)+cx+d =====(B)
Since f'(1)=4,
4=6+c (from A)
c=-2
Substituting this in B,
f(x)=g(x)+-2x+d
when x=2
f(2)=g(2)+-2(2)+d
3=9+-4+d
d=-2
Now using B,
f(x)=g(x)+-2(x)+-2
f(x)-g(x)=-2(x)-2
when x=4,
f(4)-g(4)= -8-2=-10?

what did I do wrong?Also sorry about the wrong section. I just joined 3 days ago so I thought science questions go to one section and math questions go to another section.