# Homework Help: For what values of c and d will x be differentiable for all values

1. Oct 19, 2009

### noname1

For what values of c and d will x be differentiable for all values

$$f(x)=\left\{\begin{array}{cc}cx+d,&\mbox{ if } x\leq -1\\cx^3+x+2d, & \mbox{ if } x>-1\end{array}\right.$$

i took the derivative on each side

f'(x) = c

f'(x) = 3cx²+1

c = 3cx²+1

so i get c as -(1/2)

but when i go get d i get two different values

I replace X for -1 and C for -(1/2) for the first equation i get D as-(1/2) and for the second equation i get D as (1/4) what am i doing wrong?

Last edited: Oct 19, 2009
2. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

Replace x for -1 and c for -(1/2 in which equations? You don't know what f(-1) is yet, so you can't solve for d simply by knowing what c is like that.

When you took the derivative of each side, you just assumed that f(x) was continuous at x=-1 (otherwise at least one of those values would be infinite). How can you ensure this is actually true?

3. Oct 19, 2009

### noname1

Re: differentiable

these equations:

cx + d and cx³ + x + 2d

my professor said it was continuous

4. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

So if I substitute in -1/2 and -1 for c and x I get

cx+d = 1/2+d

cx3+ x + 2d = -1/2+2d

Nothing says that both of these need to be zero, which is what it seems you did. What condition do these have to satisfy?

5. Oct 19, 2009

### noname1

Re: differentiable

i have to find the value for what c and d will x be differentiable

from what you got above the first eq gives d = -1/2 and the second gives -1/4, doest the d value have to be equal on both equations?

6. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

The equations I gave don't give d equal to anything. I don't understand where you are getting this idea that the statement cx+d = 1/2+d can possibly allow me to solve for d

7. Oct 19, 2009

### noname1

Re: differentiable

i was solving each equation independently, am i doing it wrong

8. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

What equation are you solving for d? I didn't post any equations that you can solve for d

9. Oct 19, 2009

### noname1

Re: differentiable

i was solving for the top and bottom part of

$$f(x)=\left\{\begin{array}{cc}cx+d,&\mbox{ if } x\leq -1\\cx^3+x+2d, & \mbox{ if } x>-1\end{array}\right.$$

i took the derivative of both sides

c = 3cx²+1 giving c = -(1/2) and than i was going to replace the c value and the x value which is -1 on the original equation on the left and right side

Last edited: Oct 19, 2009
10. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

What do you mean by solving for the top and bottom part of f(x)? You still haven't said exactly what equation you tried to solve, and why you tried to solve it

11. Oct 19, 2009

### noname1

Re: differentiable

the question says i have to get values for c & d for x to be differentiable on all values and i was going to replace the c value on the equation

cx+d and cx³+x+2d but seperately

for example cx+d = -(1/2)*-1 + d which equals to d = -(1/2)

but when i solve cx³+x+2d which gives -(1/2)(-1)³+2d <=> 1/2 + 2d <=> 2d = -(1/2) <=> d = -(1/4)

12. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

cx+d isn't an equation! If you know that x=-1 and c=1/2, what equation were you trying to solve?

13. Oct 19, 2009

### noname1

Re: differentiable

i updated my last post, than how am i supposed to find d? thanks for helping me and i got c by taking the derivative of both sides which is

derivative of left side = c
derivative of right side = 3cx²+1

c = 3cx²-1 <=> knowing that x is -1 from the piecewise function i get

c = 3c - 1 <=> -2c = -1 <=> c = 1/2

14. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

Ok I see
This just gives you d=d. It doesn't allow you to solve for d in any way.

Recall that for a function to be differentiable, it needs to be continuous. What equation tells you that f(x) is continuous at x=-1?

15. Oct 19, 2009

### noname1

Re: differentiable

i was replacing it from what it shows in the piece wise function, how shall i work this problem i am getting really confused

16. Oct 19, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

For f(x) to be differentiable at -1, it has to be continuous. What value of d makes it continuous? Hint: You need the two pieces to meet when x=-1

17. Oct 20, 2009

### noname1

Re: differentiable

is the c correct?

should i "connect" the left and right side of the piece wise function like this and resolve

cx+d = cx³+x+2d and than substitute c and x?

18. Oct 20, 2009

### Office_Shredder

Staff Emeritus
Re: differentiable

Looks good to me

19. Oct 20, 2009

### noname1

Re: differentiable

so d is 1 and c is -(1/2) correct?