Thread: The tangent to a graph

Here's one for ya.

Show that if is differentiable on an open interval , and if the graph of is concave upward on , then the graph of lies above all of its tangent lines on .

This is one of the problems given in my text. Although it is not required that I answer it, I feel that this problem could deepen my understanding of the second derivative test. Intuitively, I can visualize this, but I'm having trouble putting it into the language of math. Anybody got any Ideas?

Let a be any number in I

Then the tangent line at x = a is y(x) = f ' (a)(x-a) + f(a)


Let h(x) = f(x) - y(x) x > a

h '(x) = f '(x) - f ' (a) > 0 for x > a since if f '' (x) is positive then f ' (x) is increasing

Therefore h(x) is an increasing function

h(a) = f(a) - y(a) = 0

Therefore h(x) > h(a) > 0 i.e f(x) > y(x) for all x > a

Since this is true for any a in I the result follows

Originally Posted by VonNemo19

Here's one for ya.

Show that if is differentiable on an open interval , and if the graph of is concave upward on , then the graph of lies above all of its tangent lines on .

let be the equation of tangent at

then,




Let f(x)-g(x)=h(x)





Can you interpret now?

I can follow all the way up until this point. What I'm expecting to see here is an inequality stating how every . Am I wrong?

Originally Posted by malaygoel

Can you interpret now?

h(x) is an increasing function therefore f(x) > g(x) for every x in I

That is the graph of f lies above the graph of the tangent line at x0 for every x0 in I

Originally Posted by Calculus26

h(x) is an increasing function therefore f(x) > g(x) for every x in I

That is the graph of f lies above the graph of the tangent line at x0 for every x0 in I

I've got it.