Math  /  Algebra

QuestionB. What is the direction of variation of each of the following functions: (co be done by students) 1) f(x)=cosxf(x)=\cos x for 0x0 \leq x 2) g(x)x2+1x1g(x) \frac{x^{2}+1}{x-1} on its domath of definition 3) h(x)=xh(x)=\sqrt{x} on lits domain of definition

Studdy Solution
For the function h(x)=x h(x) = \sqrt{x} :
- The domain of h(x) h(x) is x0 x \geq 0 . - The derivative h(x)=12x h'(x) = \frac{1}{2\sqrt{x}} .
- Since h(x)>0 h'(x) > 0 for x>0 x > 0 , the function is increasing on its domain.
Conclusion: The function h(x)=x h(x) = \sqrt{x} is increasing for x0 x \geq 0 .
The direction of variation for each function is as follows: 1) f(x)=cosx f(x) = \cos x is decreasing on [0,π] [0, \pi] and increasing on (π,2π] (\pi, 2\pi] , repeating every 2π 2\pi . 2) g(x)=x2+1x1 g(x) = \frac{x^2 + 1}{x - 1} varies based on the sign of g(x) g'(x) around critical points 1±2 1 \pm \sqrt{2} . 3) h(x)=x h(x) = \sqrt{x} is increasing on x0 x \geq 0 .

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