Solve q#2 from d to h
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Given g(x) = (x2+1)/(x2-9)
A function is said to be increasing n certain interval, if its derivative is greater than 0 in that paticular interval. If the derivative is less than 0 in an interval then it is said to be decreasng in that particular interval.
For given function g(x), its derivative is calculated as shown below. We get g'(x) = -16x/(x2 -9)2
Here we can see that this function is not defined when x= 3, x = -3 as the denominator will be 0. Clearly we can see that for all negative values the value of g'(x) will be positive and greater than 0 and for all positive x values it will be less than 0. So we say for all negative values the function g(x) is increasing except for x = - 3 and for all positive values the function g(x) is decreasing except x = 3.
Interval where g(x) is increasing (-∞ , -3) U ( -3, 0)
Interval where g(x) is decreasing (0 , 3) U (3 , ∞)
The ordered pairs for local extrema can be found out when we equatio g'(x) = 0. Equating g'(x) to 0 we get the following as shown below. So for only x = 0 the g'(x) w...
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