f(x) = x^3 - 3x^2
= x^2(x-3)
You should be familiar with the general shape of
y = 1x^3 + .....
yours crosses at x = 3 and since you have a double root at x = 0, it touches the x-axis at x = 0
find f'(x) and f''(x) , that should be easy
when f'(x) is positive, your function increases
when f'(x) is negative, your function decreases
when f'(x) = 0 , you have "critical" points
solve f'(x) = 0 , you get the x's for those points, sub the x's back into f(x) to find the y of those points.
for points of inflection, set f''(x) = 0, find the x of the point, go back to f(x) to find the y
If f''(x) is positive, the function is concave up
if f''(x) is negative, the function is concave down
this is a pretty straightforward question, but is a good one to show all those properties
Use calculus and algebraic methods to do a complete analysis (i.e., intervals of increase and decrease, intercepts, critical points, points of inflection, and intervals of concavity) for each of the following functions and then sketch a graph of the function. (19 marks)
ƒ(x) = x3 - 3x2
2 answers
f = x^3-3x^2 = x^2(x-3)
f' = 3x^2-6x = 3x(x-2)
f" = 6x-6 = 6(x-1)
Now just apply the features you know:
f=0: x-intercepts
x=0: y-intercept
f'=0 ==> critical point
f"=0 ==> inflection
f'>0 ==> increasing
f">0 ==> concave up
f' = 3x^2-6x = 3x(x-2)
f" = 6x-6 = 6(x-1)
Now just apply the features you know:
f=0: x-intercepts
x=0: y-intercept
f'=0 ==> critical point
f"=0 ==> inflection
f'>0 ==> increasing
f">0 ==> concave up