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  • Find the values of x for which y is equal to [x (x – 2)] raised to 2 is an increasing function.

8. Find the values of x for which y = [x(x-2)]^{2}  is an increasing function.

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Given function is,
f(x)\Rightarrow y = [x(x-2)]^{2}
f^{'}(x)\Rightarrow \frac{dy}{dx} = 2[x(x-2)][(x-2)+x]
                           = 2(x^2-2x)(2x-2)
                           = 4x(x-2)(x-1)
Now,
 f^{'}(x) = 0\\ 4x(x-2)(x-1) = 0\\ x=0 , x= 2 \ and \ x = 1
So, the intervals are (-\infty,0),(0,1),(1,2) \ and \ (2,\infty)
In interval  (0,1)and \ (2,\infty) ,  f^{'}(x)> 0
Hence, f(x)\Rightarrow y = [x(x-2)]^{2} is an increasing function in the interval  (0,1)\cup (2,\infty)

Posted by

Gautam harsolia

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