Thread: The tangent to a graph

1. The tangent to a graph

Here's one for ya.

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

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?

2. 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

3. Originally Posted by VonNemo19
Here's one for ya.

Show that if $f$ is differentiable on an open interval $I$, and if the graph of $f$ is concave upward on $I$, then the graph of $f$ lies above all of its tangent lines on $I$.
let $g(x)$ be the equation of tangent at $(x_0,y_0)$

then,
$g(x)=f'(x_0)(x-x_0)+f(x_0)$

$f(x)-g(x)=f(x)-f'(x_0)(x-x_0)-f(x_0)$

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

$h(x_0)=0$

$h'(x)=f'(x)-f'(x_0)$

Can you interpret now?

4. I can follow all the way up until this point. What I'm expecting to see here is an inequality stating how every $f'(x_0). Am I wrong?

Originally Posted by malaygoel

$h'(x)=f'(x)-f'(x_0)$

Can you interpret now?

5. 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

6. 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.