Question

If f(x)=x/[sqrt(1+x^2)],
then (fofof)(x)=

Answer 1

hi friend,

f(x)=x/√(1+x²)

→f(f(x))=[ x/√(1+x²)]/√(1+(x/√(1+x²))²)

→[x/√(1+x²)]/√(1+(x²/1+x²))

→[x/√(1+x²)]/√(1+2x²/1+x²)

→x/√(1+2x²)

now ,f(f(f(x)))=[x/√(1+x²)]/√(1+2(x/√(1+x²)²)

→[x/√(1+x²)]/√(1+(2x²/1+x²))

→[x/√(1+x²)]/√(1+3x²/1+x²)

→x/√(1+3x²)

I hope this will help u :)

Answer 2

Answer:

f(x)=x/V(1+x2)

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)+[x//(1+x*}J/v(1+(x®/1+x³))

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)+[x//(1+x*}J/v(1+(x®/1+x³))[x/v/(1+x?)]//(1+2x?/1+x²)

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)+[x//(1+x*}J/v(1+(x®/1+x³))[x/v/(1+x?)]//(1+2x?/1+x²)+x/-/(1+2x2)

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)+[x//(1+x*}J/v(1+(x®/1+x³))[x/v/(1+x?)]//(1+2x?/1+x²)+x/-/(1+2x2)now ,f(f(f(x)))=[x/(1+x²)]/v(1+2(x/v(1+x²)?)

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)+[x//(1+x*}J/v(1+(x®/1+x³))[x/v/(1+x?)]//(1+2x?/1+x²)+x/-/(1+2x2)now ,f(f(f(x)))=[x/(1+x²)]/v(1+2(x/v(1+x²)?)+[x/V(1+x³)J/v(1+(2x?/1+x*))

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)+[x//(1+x*}J/v(1+(x®/1+x³))[x/v/(1+x?)]//(1+2x?/1+x²)+x/-/(1+2x2)now ,f(f(f(x)))=[x/(1+x²)]/v(1+2(x/v(1+x²)?)+[x/V(1+x³)J/v(1+(2x?/1+x*))+[x//(1+x?)]//(1+3x²/1+x²)

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)+[x//(1+x*}J/v(1+(x®/1+x³))[x/v/(1+x?)]//(1+2x?/1+x²)+x/-/(1+2x2)now ,f(f(f(x)))=[x/(1+x²)]/v(1+2(x/v(1+x²)?)+[x/V(1+x³)J/v(1+(2x?/1+x*))+[x//(1+x?)]//(1+3x²/1+x²)+x/V(1+3x?)

f(x)=x/V(1+x2)+f({f{x))=[ x/v(1+x³)]/v(1+(x/v(1+x³))^)+[x//(1+x*}J/v(1+(x®/1+x³))[x/v/(1+x?)]//(1+2x?/1+x²)+x/-/(1+2x2)now ,f(f(f(x)))=[x/(1+x²)]/v(1+2(x/v(1+x²)?)+[x/V(1+x³)J/v(1+(2x?/1+x*))+[x//(1+x?)]//(1+3x²/1+x²)+x/V(1+3x?)Step-by-step explanation:

Hope it will help