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.
How do I solve it analytically? I'm still confused.
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.
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.
To determine the statements that are NOT true for the graph of f(x) = (x+1)(x-3)^2, let's analyze each option separately.
A. "f has a relative maximum at (⅓, 256/27)"
To find the relative extrema, you should set the derivative f'(x) = 0 and solve for x. In this case, you stated that f'(x) = 3x^2 - 10x + 3. So, to find the critical points, we need to solve the equation: 3x^2 - 10x + 3 = 0.
However, since you mentioned your critical numbers are weird fractions, let's double-check your derivative calculation. The correct derivative of f(x) = (x+1)(x-3)^2 should be:
f'(x) = (2(x-3)(x+1)) + ((x-3)^2)
Let's expand this:
f'(x) = 2(x^2 - 2x - 3) + (x^2 - 6x + 9)
= 2x^2 - 4x - 6 + x^2 - 6x + 9
= 3x^2 - 10x + 3
Now, to find the critical points, we solve:
3x^2 - 10x + 3 = 0
Using any method of solving quadratic equations (factoring, quadratic formula, completing the square), we find that x = (5 ± √7)/3. These are the critical numbers, not the weird fractions you mentioned.
Based on the correct critical points, option A is true. Therefore, it is NOT the statement that is false.
B. "f has a point of inflection at (3, 0)"
To find inflection points, we need to identify where the concavity of the function changes. This occurs when the second derivative, f''(x), changes sign. You calculated f''(x) = 6x - 10, which is correct.
Setting f''(x) = 0 and solving for x, we get:
6x - 10 = 0
6x = 10
x = 10/6
x = 5/3
This is a critical point. However, it does not lie on the given function f(x) = (x+1)(x-3)^2, so it cannot be a point of inflection. Therefore, option B is NOT true.
C. "f has an intercept at (3, 0)"
To find intercepts, we look for points where the graph crosses the x-axis (y = 0) or the y-axis (x = 0). In this case, the point (3, 0) is indeed an x-intercept since f(3) = 0. Hence, this statement is true.
D. "f has a relative minimum at (3, 0)"
To determine relative extrema, we need to look for points where the derivative changes sign. However, we already found that the critical point at x = 5/3 is not on the original function f(x) = (x+1)(x-3)^2. Therefore, there is no relative minimum at (3, 0), and this statement is NOT true.
Considering the analysis above, we conclude that the statement NOT true of the graph f(x) = (x+1)(x-3)^2 is option D.