If f(x) = {(a+x)/(1+x)}^a+1+2x then the value of f’(0) is ?
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f(f(x))=
x+1
ax
+1
a
x+1
ax
=
x+1
ax+x+1
x+1
a
2
x
=
ax+x+1
a
2
x
Since, f(f(x))=x, we have,
ax+x+1
a
2
x
=x.
Simplifying the equation we get,
a
2
x=(a+1)x
2
+x
∴(a+1)x
2
+(1−a
2
)x=0
or (a+1)x(x+1−a)=0
Hence the only possible value is a=−1
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