# Thread: tricky tangent qroblem :)

1. ## tricky tangent qroblem :)

Suppose g(x) has a second derivative g''(x) at all points of an interval I. Prove that if the tangent to the curve y=g(x) at a in I meets the curve again at some other point b>a in I then g''(x)=0 somewhere between a and b...

I seriously dont know where to begin

2. Originally Posted by Khonics89
Suppose g(x) has a second derivative g''(x) at all points of an interval I. Prove that if the tangent to the curve y=g(x) at a in I meets the curve again at some other point b>a in I then g''(x)=0 somewhere between a and b...

I seriously dont know where to begin

The phrase 'meets the curve again' refers to inflection. So, I would start by understanding that wherever the tangent line to a graph crosses the graph, $\displaystyle f''(x)=0$. There is a proof of this fact here:

Second derivative test - Wikipedia, the free encyclopedia

3. Hey von-

I'm sorry but I still dont understand what your trying to tell me

can you explain wat exact is the question asking me.

4. Originally Posted by Khonics89
Suppose g(x) has a second derivative g''(x) at all points of an interval I. Prove that if the tangent to the curve y=g(x) at a in I meets the curve again at some other point b>a in I then g''(x)=0 somewhere between a and b...

I seriously dont know where to begin
first note that $\displaystyle \frac{g(b)-g(a)}{b-a}=g'(a),$ because they are both equal to the slope of the tangent line to the curve at $\displaystyle x=a.$ call this (1). define $\displaystyle f(x)=g(x)-\frac{g(b)-g(a)}{b-a}(x-a)-g(a), \ x \in I.$

see that $\displaystyle f(a)=f(b).$ thus by Rolle's theorem there exists $\displaystyle a<c<b$ such that $\displaystyle g'(c)-\frac{g(b)-g(a)}{b-a}=f'(c)=0,$ i.e. $\displaystyle g'(c)=\frac{g(b)-g(a)}{b-a}.$ therefore $\displaystyle g'(a)=g'(c),$ by (1). again, applying Rolle's

theorem, gives us that $\displaystyle g''(d)=0,$ for some $\displaystyle a<d<c,$ and we're done.