Show that the function f (x) = x^{2} is strictly increasing function on [0,∞).

### Asked by Topperlearning User | 6th Aug, 2014, 08:12: AM

Let *x*_{1}, *x*_{2} [0,¥) such that *x*_{1} < *x*_{2}. Then,

*x*_{1} < *x*_{2} => *x _{1}^{2} < x_{1}x_{2..........}*(

*i*) [Multiplying both sides by

*x*

_{1}]

and, *x*_{1} < *x*_{2 }=> *x _{1}x_{2}<x_{2}^{2}* ............ (

*ii*) [Multiplying both sides by

*x*

_{2}]

From (*i*) and (*ii*), we get

*x*_{1} < *x*_{2} => *x*_{1}^{2} < *x*_{2}^{2} => *f* (*x*_{1}) < *f* (*x*_{2})

Thus, *x*_{1} < *x*_{2} => *f* (*x*_{1}) < *f* (*x*_{2}) for all *x*_{1}, *x*_{2} [0, ¥)

Hence, *f* (*x*) is strictly increasing function on [0, ¥)

### Answered by | 6th Aug, 2014, 10:12: AM

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