Doug C. answered • 05/22/21
Math Tutor with Reputation to make difficult concepts understandable
The derivative does show 2x2-30x in the numerator, but there is also (x2+2x-15)2 in the denominator.
Setting that equal to zero, you can think of multiplying both sides by that denominator which will result in:
2x2 - 30x = 0 (in other words a fraction can only equal zero if the numerator equals zero)
Now solve by factoring to get the critical numbers:
2x(x-15) = 0, so,
x=0 or x = 15.
Those are the points where the graph of the original function has horizontal tangent lines. Use the first derivative test to determine if those critical numbers generate relative maximum or minimum (or neither).
Girlie S.
hey i appreciate this! i did the first derivative test and got 0 for both, does this mean that it does not have a min or max? or the max is (0,0) ?05/22/21
Doug C.
The critical numbers are 0 and 15. If you factor the denominator you will get (x+5)(x-3) which means there are vertical asymptotes at x=-5 and x = 3. Here is an updated graph: desmos.com/calculator/owl4sgixxt If you look at the last rows of the graph you will see that the derivative, g(x) on this graph, is evaluated at various values in the intervals created by the asymptotes and the critical numbers. For example, g(-4) is positive (so original function is increasing) and g(1) is negative (original function is decreasing). That means at (0,0) there is a relative maximum--that is the first derivative test. Similarly, g(14) is negative (original function decreasing) and g(16) is positive (original function is now increasing). That means relative min at (15, .9375).05/22/21
Doug C.
desmos.com/calculator/oxpiajhy5g Forgot to include this link in the answer.05/22/21