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

[noname1]

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

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

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?

thanks in advance

 

[Office_Shredder]

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?

 

[noname1]

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?

these equations:

cx + d and cx³ + x + 2d

my professor said it was continuous

 

[Office_Shredder]

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?

 

[noname1]

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?

 

[Office_Shredder]

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

 

[noname1]

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

 

[Office_Shredder]

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

 

[noname1]

i was solving for the top and bottom part of

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

 

[Office_Shredder]

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

 

[noname1]

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 separately

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)

 

[Office_Shredder]

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

 

[noname1]

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

 

[Office_Shredder]

Ok I see

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

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?

 

[noname1]

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

 

[Office_Shredder]

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

 

[noname1]

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?

 

[Office_Shredder]

Looks good to me

 

[noname1]

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