Determine the value of the constant that makes f'(0) exists

[Marcia]

Let f(x) = cSinx x less than 0
5x^2 = 5x x is greater than or equal to 0


This is the problem. I am asked to Determine the value of the constant c that makes f'(0) exist.

Any help would be appreciated.
Thanks.

 

[Unco]

hmmm... I could be horribly mistaken, but f'(x) = c.cos(x)... and this will be defined for all values of x. Do you mean f'(0) is not to equal zero?

 

[Marcia]

Puzzled myself

The question reads:

Let f(x) = { csin x x is less than 0
5x^2 + 5x x is greater than or equal to 0

Determine the value of the constant c that makes f'(0) exists.

That is all there is for the question.

 

[soroban]

Re: Determine the value of the constant that makes f'(0) exi

Hello, Marcia!

There are obvious typos in the problem.

Let f(x) = c·sinx . . . . x < 0
. . . . . . . 5x² = 5x . . x <u>></u> 0
. . . . . . . \______/
. . . . . . . . . . ?
Determine the value of the constant c that makes f'(0) exist.

. . . . . . . . . . . . . . . . . . . . . . . . . . . . | c·sin x . . . . x < 0
I'll assume that we have: . f(x). = . |
. . . . . . . . . . . . . . . . . . . . . . . . . . . . | 5x² + 5x . . x <u>></u> 0

We must ensure that the values of f '(0) are equal "from both sides".

From the left: .f '(x) .= .c·cos x
. . When x = 0: .f '(0) .= .c·cos(0) .= .c

From the right: .f '(x) = 10x + 5
. . When x = 0: .f '(0) .= .10(0) + 5 .= .5

Since these derivative must be equal: .c = 5

 

[Marcia]

Thank you.

That is what I had for the answer.