106Q2. Consider an odd function f(x)
then f(x) = -f(-x)
f'(x) = -f'(-x)(-1)
f'(x)= f'(-x)
Consider an even func f(x)
=> f(x) = f(-x)
=> f'(x) = -f'(-x)
=> f'(x) is odd
clearing all my doubts today ...
Pick the correct options from the following for the below question , and state reasons for the false ones
Q1] If g(x) is an inverse function of f(x) then
a] g(x)=f-1(x)
b] f(g(x)) =x
c] f(x) = g-1(x) [ i.e if g is inverse of f then f is inverse of g ]
d] g-1[f-1(x)] = x
e] f must be one one onto
f] g'(x) = \frac{1}{f'(f^{-1}(x))}
Q2] in arihant it is given that :
derivative of an odd function is always an even function, and derivative of an even function is always an odd function
is it correct ?? i doubt it ....waiting for your views
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6 Answers
106106Q1. for inverse to exist (e) is a must
g(f(x)) = x (g(x) is inverse of f(x)) .. (i) (f inverse of f(x) = x)
f(g(x)) = x ... (ii) (f of f inverse x = x)
=>g-1(g(f(x))) = g-1(x)
=> f(x) = g-1(x)
similarly
g(x)=f-1(x) (taking inverse of f(x) on both sides of (ii))
g-1[f-1(x)] = x can be proved by taking g inverse of the expression derived in the just previous step
thinking abt (f) gotta go
106As f(g(x)) = x
=> f'(g(x))*g'(x) = 1
=> g'(x) = 1/f'(g(x)) = 1/f'(f-1(x))
so all correct
23u sure na all correct ??
and wat about c ???
in arihant he gives dat
if fog = gof then only f(x) = g-1(x)
and f-1(x) = g(x) hold simultaneously ...
1the derivative of an odd function is even and vice versa only if the function is not constant