IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

kishan12

The domain of the derative of the function
f(x) = tan(inverse)x if mod of x <or=1
1/2 (mod of x -1) if mod of x > 1

Plzzz answer
salutes gaurranteed

neeraj_agarwal_1990

Making the graph of the given function wud make the domain clear....

this function would be discontinuous at x=1 and -1...and it is smooth everywhere ....so domain is R-{-1,1}

jus_look

f(x)=tan(inverse)x; x lies b/w -1 n 1(including both)
=(1-x)/2;x less than -1
=(x-1)/2;x greater than 1
so deriv of f(x)=1/(1+x square)
=-1/2
=1/2 limits as same above
but its discont at -1 n 1
so domain is r-{-1,1}

rate me if m right.....

johri_anshuman

f(x)=tan^-1x   |x|<=1
      (1/2)|x+1|   |x|>1

so f(x) can be redefined as

f(x)=tan^-1x   x

[-1,1]
-(1/2)(x+1) x

(-infinity,-1)
(1/2)(x+1) x

(1,infinity)

So the graph of the function will be as shown.

clearly from the graph f(x) is not differentiable at x=-1,1

hence domain of derivative of f(x) is R-{-1,1}

spideyunlimited

f'(x) = 1 / 1 + x^2 |x|<=1
1/2 x>1
-1/2 x< -1

there fore only factor for domain of derivative is denominator in 1 / 1 + x^2 and where |x|<=1

thus, x can have values between - 1 to 1.
thus domain of f(x) = [-1 , +1]

*PLZ RATE*

joyfrancis

f'(x) would be

1/1+x^2 ........for x

[-1,1]

1/2...........for x

(1,inf)

-1/2..........for x

(-inf,1)

Clearly the function is discontinuous at x=-1 and x=1, so the domain of f'x is

R-{-1,1}

spideyunlimited

fn is discontinuous ar x= 1 also

joyfrancis

oh ya!, i edited it. Silly Mistake!

pavan_puthr

Well for the first function.....

ie f(x) = tan^-1(x)

f '(x) = ¹/_(1+x²)

from this expression it can be deduced that f '(x) does not exist for

1

But in the question it is given |x|

1 ie -1

x

1

so the domain of the derivativebecomes

-1 < x < 1

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Now for the second question

ie f (x) = ¹/₂ |x - 1|

it can be derived that (^d/_dx)(|x|) = ^x/_|x| (*)

so, for our function f '(x) = ^{(x - 1)}/_{|x - 1|}

obviously f '(x) is undefined for x=1

but from question x > 1

so the domain of the derivative is also x > 1 or

x

( 1,

)

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(*) nudge me if u want the proof of f '(x) of f (x)=|x|

______________________________________________________________ Rate me okay

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