Thread: [CALC 1]: Find a set of real numbers a, b, and c for which this piecewise-defined....

1. [CALC 1]: Find a set of real numbers a, b, and c for which this piecewise-defined....

Find a set of real numbers a, b, and c for which this piecewise-defined function is (everywhere) differentiable:

f(x)= tan-1x, x≤ 0
f(x)= ax2 + bx + c, 0 < x < 2
f(x)= x3- 1/4x2 + 5, x ≥ 2

so i plugged in x for the numbers and got:
tan-1(0)= 0
a(0)2 + b(0) + c = 0
a(2)2 + b(2) + c= 4a+2b+0
(2)3+ 1/4(2)2 + 5= 12

tan-1(0)= a(0)2 + b(0) + c
c= 0

4a+2b= 12
-->2a+ b= 6

i didn't know how to do the rest from here.
please show all steps. thank you!

2. Re: [CALC 1]: Find a set of real numbers a, b, and c for which this piecewise-defined

Originally Posted by krutikahegde
Find a set of real numbers a, b, and c for which this piecewise-defined function is (everywhere) differentiable:

f(x)= tan-1x, x≤ 0
f(x)= ax2 + bx + c, 0 < x < 2
f(x)= x3- 1/4x2 + 5, x ≥ 2

so i plugged in x for the numbers and got:
tan-1(0)= 0
a(0)2 + b(0) + c = 0
a(2)2 + b(2) + c= 4a+2b+0
(2)3+ 1/4(2)2 + 5= 12

tan-1(0)= a(0)2 + b(0) + c
c= 0

4a+2b= 12
-->2a+ b= 6

i didn't know how to do the rest from here.
please show all steps. thank you!

So far so good. For a function to be differentiable we require that the first derivative is continuous also.

Can you finish from here?

-Dan

ok

4. Re: [CALC 1]: Find a set of real numbers a, b, and c for which this piecewise-defined

One mistake I see ...

$f(2)=2^3+\dfrac{1}{4} \cdot 2^2 + 5 = 8 + 1 + 5 = 14$

$4a+2b=14 \implies 2a+b=7$

You already have $c=0$, now, as stated by topsquark, $f'(x)$ also needs to be continuous ...

$f'(x)=\dfrac{1}{1+x^2} \,\,\, x \le 0$

$f'(x) = 2ax+b \,\,\, 0 < x <2$

Get the idea?

5. Re: [CALC 1]: Find a set of real numbers a, b, and c for which this piecewise-defined

thats not a mistake. -1/4 not plus 1/4
so it would be 2^3 - 1/4 (2)^2 + 5 = 12....

i meant 2^3 - 1/4 on the equation. not + 1/4

6. Re: [CALC 1]: Find a set of real numbers a, b, and c for which this piecewise-defined

Originally Posted by topsquark
So far so good. For a function to be differentiable we require that the first derivative is continuous also.

Can you finish from here?

-Dan
No, we do NOT require that the derivative be continuous in order that the function be differentiable. But it is true that even if the derivative is not continuous, it must still satisfy the "intermediate value property" (that can be proved by applying the "mean value property" to the original function) and that implies that if the two "one sided limits" at a point exist, then they must be equal and equal to the derivative so the derivative is continuous there- and this applies to the problems given here. But it is possible for the derivative of a given function to exist at every point but one or both of the one sided limits does not exist at a given point and the derivative not be continuous there.

(The condition that the derivative itself be continuous is often given as "a function is continuously differentiable.)