# F(x)={sinx for x<22/7 {mx+b for x>22/7

1. Feb 23, 2005

### kidia

Hi any idea on this:
Find all the values of the constants b and m for which the function.

f(x)={sinx for x<22/7
{mx+b for x>22/7

I will appreciate for any help

2. Feb 23, 2005

### HallsofIvy

Rule 1: Copy the problem correctly! The way you've stated it, you can't find m and b. There exist functions having those values for all m and b! Did you mean "for which the function,..., IS DIFFERENTIABLE at x= 22/7"?

In order to be differentiable at a point, the function must first be continuous there.
Strictly speaking, the function you gave is not even defined at 22/7 and so can't be continuous there.

I am going to assume that what you REALLY mean is: "Find all values of the constants b and m for which the function
{sin x for x<= 22/7
f(x)= { mx+ b for x> 22/7

is differentiable." (Since f is clearly differentiable every where other than 22/7 the crucial part is that it be differentiable at 22/7).

In order to be differentiable, it must first be continuous: the limit of f(x) at 22/7 FROM THE LEFT (x< 22/7) is sin(22/7) (And, once again: 22/7 is ALMOST pi. Did your problem really have 22/7 rather than pi). The limit of f(x) at 22/y FROM THE RIGHT is
22m/7+ b so we must have 22m/7+ b= sin(22/7) (which is close to 0).

The derivative of f, for x< 22/7 is cos(x). The derivative of f for x> 22/7 is m. While it is not necessary that derivatives be continuous, they do satisfy the "intermediate value problem" so if f is differentiable at 22/7, and since the derivative functions to the left and right of 22/7 are continuous, they must "match up" there: m= cos(22/7) (close to -1). You can put that into the other equation to solve for b. (In fact, mx+ b is simply the tangent line to sin(x) at x= 22/7.)

3. Feb 23, 2005

### kidia

sorry,sorry,sorry is true the one sent to me copied the problem wrong is like this

Find all the values of the constants b and m for which the function.

f(x)={sin x for x<pi
{mx+b for x>=pi
is a)continuous at x=pi b)is differentiable at x=pi

4. Feb 24, 2005

### HallsofIvy

Okay- in order fror the function to be continuous at pi, we must have
$$m\pi+ b= sin(\pi)= 0$$
In order for the function to be differentiable at pi, we must also have
$$m= cos(\pi)= -1$$.

5. Feb 24, 2005

### kidia

Thanx I understand u