Question: find the approximate location of all relative extrema f(x) = 0.1x^3-15x^2+96x+91.
f = 0.1x^3-15x^2+96x+91
Turning points at f'(x) = 0
f'(x) = 0.3x^2 - 30x + 96 = 0
Using the quadratic formula to solve the above quadratic
x = (30 ± √(30^2 - 4*0.3*96))/(2*0.3)
Solutions are: x1 = 3.309529880, x2 = 96.69047012
Type of extrema using sign of f''(x)
f''(x) = 0.6x - 30
1st turning point - x1 = 3.309529880
f''(x1) = f''(x = 3.309529880)= 0.6*3.309529880 - 30 = -28.014
SInce f''() is negative, then hte slope is decreasing, so the TP is at a maximum.
2nd turning point - x2 = 96.69047012
f''(x2) = f''(x = 96.69047012)= 0.6*96.69047012 - 30 = 28.014
SInce f''() is positive, then the slope is increasing, so the TP is at a minimum.
Answer: The turning points are at x1 = 3.3 (a maximum) and x2 = 96.7 (a minimum)