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6) Find the intervals in which the following functions are strictly increasing or
decreasing 

b) 10 - 6x - 2x^2

Answers (1)

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Given function is,
f(x) = 10 - 6x - 2x^2
f^{'}(x) = -6 - 4x
Now,
f^{'}(x) = 0
6+4x= 0
x= -\frac{3}{2}

So, the  range is (-\infty , -\frac{3}{2}) \ and \ (-\frac{3}{2},\infty)
In interval (-\infty , -\frac{3}{2})  ,  f^{'}(x) = -6 - 4x is +ve
Hence, f(x) = 10 - 6x - 2x^2 is strictly increasing in the interval  (-\infty , -\frac{3}{2})
In interval ( -\frac{3}{2},\infty)  , f^{'}(x) = -6 - 4x is -ve
Hence, f(x) = 10 - 6x - 2x^2 is strictly decreasing in interval  ( -\frac{3}{2},\infty)

Posted by

Gautam harsolia

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