Question

determine whether function f is invertible: f'(x)=x^2+1 f'(x)=x^2-1 f'(x)=sinx f'(x)=pi/2+tan^-1(x)

determine whether function f is invertible: f'(x)=x^2+1 f'(x)=x^2-1 f'(x)=sinx f'(x)=pi/2+tan^-1(x)

Math   Calculus   
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Answer #1

if f'(x) = 0 , and the critical point is point of local maximum and minimum than function is not one-one. Hence, it is not invertible. For invertible function must be one - one and onto .

1>

therefore , f(x) is always increasing function.

f(x) is invertible

2>

, x = 1 is point of minimum

, x = -1 is point of maximum

Function is not one-one, hence not invertible

3>

this is periodic function, with critical point at x =

f(x) = cos(x) +C, this is also periodic function. Periodic function is not one-one. Therefore, it is not invertible

4>

because range of

therefore f(x) is always increasing function, making it one-one onto, hence it is invertible function.

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