Question

show that the function f: R R defined by f(x)= x^2+1 neither one one nor onto​

Answer 1

$\huge\boxed{Answer}$

By a little examination, we can see that:-

For all x>0, f(x)∈(0, 1)

For x=0, f(x)=0

For all x<0, f(x)∈(−1, 0)

So, clearly f(x) is not onto, as its range belongs to only (−1, 1), and not (-R, R).

Now, f '(x)= (11+x2)ddx(x) + xddx(11+x2)         =11+x2+x{−2x(1+x2)2}         =11+x2 − 2x2(1+x2)2

                   =1+x2−2x2(1+x2)2=1−x2(1+x2)2

So, f '(x)=0 at x=1.

This means, that the slope of f(x) changes sign before and after x=1.

That is, there exists some x¹, x²and x³ for which f(x¹) =f(x²)

So, f(x) is not one-one.

$\huge\mathfrak\red{itz\:jyotsana☺}$