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Hi here's the question:

It is given that f(x)=2x^3+3x^2+1.
a) Find the derivative f'(x), factorising your answer.
b) Hence find the coordinates of the stationary points on the curve and determine their nature.
c) Find the values of x for which f(x) is an increasing function.
I have worked out all the answers for these apart from c) because I don't understand the 'range of values of x'. Do I need to sketch a graph of the curve to find some inequalities?
Thanks a lot.

Answer
Hi Jordan,
The derivative of f(x) = 2x + 3x + 1 is given by f'(x) = 6x + 6x = 6x(x + 1) and the stationary points for f(x) are gotten from f'(x) = 0 as you would have already done.
Now, a function f(x) is increasing in an interval (a, b) if for all a < b, f(a) ≤ f(b). This just means that in that interval f(x) increases as x increases. For a function f(x) whose derivative is defined at all points in the interval (as with the function we're dealing with here) the function is increasing when f'(x) > 0. Again, we know that the derivative is the rate of change of the function and so this simply means that function is increasing when we have a positive rate, as we would expect.
Anyway, f(x) = 2x + 3x + 1 is increasing for all f'(x) = 6x(x + 1) > 0 with the solution as the two intervals x < -1 and x > 0.
The way to solve the inequality is to consider that 6x(x + 1) = 0 at x = -1 and x = 0 and then divide the entire number line (without -1 and 0) into the three intervals x < -1, -1 < x < 0 and x > 0. You then take any number in each interval to test it. For instance, in the first interval we could take x = -5 and then find that f'(-5) = 6(-5)(-5 + 1) = 6(-5)(-4) = 120 which is > 0 and so the function is increasing in that interval.

Regards

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