Find a & b such that f(x) is differentiable everywhere?

Question

f(x) = x^2 + 2x - 3 if x is less than or equal to 1
ax + b if x is greater than 1

find a and b

Answer 1

both parts of the piecewise function are differentiable on their domain.

This means you only have to worry about differentiability the point where they meet at x = 1

To even have a chance of being differentiable at this point the the function must be continuous at this point (in other words, the function for x<= 1 must equal the function for x> 1 at x =1)

f(x) = x^2 +2x -3
f(1) = 1+2-3 = 0

so ax+b must also = 0 at x = 1

a + b = 0

This will allow the function to be continuous at x = 1, which will give it the possibility of being differentiable..

in order for it to be differentiable. the slope of both parts of the piecewise function must be the same at x = 1 (otherwise you'd have a sharp corner)

f(x) = x^2 +2x - 3
f'(x) = 2x +2
f(1) = (2)(1) + 2 = 4

so a = 4

referring back to a+b = 0

4+b = 0
b = -4

a= 4 b =-4

Answer 2

P.S. I KNOW SOME OF THE OTHER ANSWERERS HAVE GOT THE ANSWER AS a=4 & b=4; but they have to note that they got the answer because they assumed the function to be continuous at x=1. However, that is not necessarily correct. A differentiable function DOES NOT have to be continuous at any point - only the left-hand and right-hand derivatives have to be equal. You may wanna check it up with someone who is experienced at math.

Anyway, here's the answer for your problem:

Since f(x) = x^2 + 2x - 3 is a quadratic equation, it is differentiable everywhere in it's range, i.e. less than or equal to 1.

Also, since f(x) = ax + b is a linear equation, it is also differentiable everywhere in it's range, i.e. greater than 1.

So, you really only have to check whether it is differentiable at the point of intersection of these two curves, i.e. at x=1.

Left hand limit of derivative:

lim(h->0) f(x-h) - f(x) / (-h)

lim(h->0) f(1-h) - f(1) / (-h)

lim(h->0) { (1-h)^2 +2(1-h) - 3 - 1^2 - 2(1) + 3 } / (-h)

lim(h->0) { (1-h)^2 +2(1-h) - 3} / (-h)

lim(h->0) { 1 - 2h + h^2 + 2 - 2h - 3 } / (-h)

lim(h->0) { h^2 - 4h } / (-h)

lim(h->0) { h - 4 } / (-1)

= 4



Right hand limit of derivative

lim(h->0) f(x+h) - f(x) / (+h)

lim(h->0) f(1+h) - f(1) / (+h)

lim(h->0) { a(1+h) + b - a(1) - b } / h

lim(h->0) { a + ah - a } / h

= a



If our function has to be differentiable at x=1, then the left & right limits should be equal. Thus, a = 4.

a=4 is the answer, b can take any value.

Answer 3

lim f(x) x=>1 from the left is 0
Must be the same from tthe right.So a+b=0
The derivatives must be equal
f´(x) = 2x+2 = 4 at x=1 So a=4 and b=-4

Answer 4

If you substitute 1 into x for both expressions, both answers should be the same. Let this be your first equation.

Now differentiate both, and use 1 for x again, and you should again get the same. Let this statement be your second equation.

Solve the system.

Answer 5

Find a and b so that.

A) f(1) = a*1 + b
B) f'(1) = a (which is the derivative of ax + b at 1)