Check out the graph, and scroll down to information on the extrema at
http://www.wolframalpha.com/input/?i=%28x%2B1%29%28x-3%29^2
I think you are doing things right.
Which statement is NOT true of the graph f(x)=(x+1)(x-3)^2?
A. f has a relative maximum at (⅓, 256/27)
B. f has a point of inflection at (3, 0)
C. f has an intercept at (3, 0)
D. f has a relative minimum at (3, 0)
E. none of these
For f'(x) I got 3x^2-10x+3
For f"(x) I got 6x-10
My critical numbers and inflection points are these weird fractions so I think I might be doing something wrong.
3 answers
How do I solve it analytically? I'm still confused.
well, you just find where
f'=0 and f"≠0 for extrema
f"=0 for inflection points
f' = (3x-1)(x-3)
so, f'=0 at x = 3, 1/3
f"(1/3) < 0, so f(1/3) is a maximum
f"(3) > 0, so f(3) is a minimum
f"=0 at x = 5/3, so there's an inflection point there.
You can surely see that B and D cannot ever both be true. You cannot have an inflection point and an extremum at the same place.
Just review carefully the relevant examples in your text.
f'=0 and f"≠0 for extrema
f"=0 for inflection points
f' = (3x-1)(x-3)
so, f'=0 at x = 3, 1/3
f"(1/3) < 0, so f(1/3) is a maximum
f"(3) > 0, so f(3) is a minimum
f"=0 at x = 5/3, so there's an inflection point there.
You can surely see that B and D cannot ever both be true. You cannot have an inflection point and an extremum at the same place.
Just review carefully the relevant examples in your text.
3 answers