- #1
dharavsolanki
- 77
- 0
Homework Statement
Given P(x) = x4 + ax3 + bx2 + cx + d such that x = 0 is the only real root of P'(x) = 0. If P(–1) < P(1), then in the interval [–1, 1]
(1) P(–1) is not minimum but P(1) is the maximum of P
(2) P(–1) is minimum but P(1) is not the maximum of P
(3) Neither P(–1) is the minimum nor P(1) is the maximum of P
(4) P(–1) is the minimum and P(1) is the maximum of P
Homework Equations
The answer is 1.
The Attempt at a Solution
- P(x) = x4 + ax3 + bx2 + cx + d
- P′ (x) = 4x3 + 3ax2 + 2bx + c
- P′ (0) = 0
- c = 0
- P′ (x) = 0 only at x = 0
- P′ (x) is a cubic polynomial changing its sign from (–)ve to (+)ve and passing through O.
- P′ (x) < 0 ∀ x < 0
- P′ (x) > 0 ∀ x > 0
- graph of P(x) is upward concave, where P′ (x) = 0
Now P(–1) < P(1)
⇒ P(–1) cannot be minimum in [–1, 1] as minima in this interval is at x = 0.
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The question is solved. That is not the problem. What I am wondering about right now is that such a logical question is too precious to be let go of without a discussion. Do you have any comments regarding either the geometrical interpretation or logical deduction of the answer?
For me, the questions are getting monotonous by the day and I have almost lost the thrill of learning because now the questions are just too clear. However, I am sure you will have a lot to discuss about it, right?
