A circle centered at the origin contains the point
1 answer:
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9514 1404 393
Answer:
- graph is shown below
- absolute max and min do not exist
- local max: 0 at x=0
- local min: -500/27 ≈ -18.519 at x=10/3
Step-by-step explanation:
The function is odd degree so has no absolute maximum or minimum. It factors as ...
g(x) = x^2(x -5)
so has zeros at x=0 (multiplicity 2, meaning this is a local maximum*) and x=5.
Differentiating, we find the derivative of g(x) is zero at x = 0 and x = 10/3.
g'(x) = 3x^2 -10x = x(3x -10) ⇒ x=0 and x=10/3 are critical points
The value of g(10/3) is a local minimum. That value is ...
g(10/3) = (10/3)^2((10-15)/3) = -500/27 ≈ -18.519
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The local maximum is (0, 0); the local minimum is (10/3, -500/27). The graph is shown below.
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* When a root has even multiplicity, the graph does not cross the x-axis. That means the root corresponds to a local extremum. Since this is the left-most root of an odd-degree function with a positive leading coefficient, it is a local <em>maximum</em>. (The function is <em>increasing</em> left of the left-most turning point.)
