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Calculus/Derivatives at 0, intervals where the function is increasing/decreasing

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Expert: - 2/8/2010


Question
h(x)= (25x^2+15x-9)(5x+3)^2

I need to find the derivative of the function and simplify the result to find the points where the derivative is 0 and the intervals of the function is increasing/decreasing. I can typically get to the derivative function, however I am having difficulty simplifying the function enough to get anywhere on the problem.

Answer
The function is h(x) = (25x² + 15x - 9)(25x² + 30x+ 9).
The derivative would be found by the product rule on the two functions shown.

That would be (25x² + 15x - 9)(50x + 30) + (50x + 15)(25x² + 30x + 9).

Multiplying it out gives 1250x³ + 1500x² - 270  +  1250x³ + 1875x² + 900x + 135.

Combining like terms gives 2500x³ + 3375x² + 900x - 135.
I have found one of the roots to be -0.6, so the equation breaks down to
(x+0.6)(2500x²+1875x-225), whichcan be seen to be 25(x+0.6)(100x²+75x-9).

Now the quadratic formula used on 100x²+75x-9, as b²-4ac, 5,625 - 400*(-9) = 9,225.
Note that is divisible by 25, so the √9225 = 5√369.
Next, note 360 = 9*31, so this is really 15√41.

Using the -b and the 2a, we get (-75 ± 15√41)/200.

Evauluate the derivative at points in between and see whether it is positive or negative.
If the derivative is positive, the function is increasing.
If the derivative is negative, the function is decreasing.  

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