Let f ( x ) = ( x − a ) ( x − b ) 3 ( x − c ) 5 ( x − d ) 7 , where a , b , c , d real numbers with a < b < c < d .
Consider the derivative of f ( x ) , f ′ ( x ) . Find the number of distinct real roots of f ′ ( x ) = 0 .
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well @Otto Bretscher sir, I have found a relatively more analytical method.. f ( x ) = ( x − a ) 1 ( x − b ) 3 ( x − c ) 5 ( x − d ) 7 f / ( x ) = ( x − b ) 2 ( x − c ) 4 ( x − d ) 6 { ( x − b ) 1 ( x − c ) 1 ( x − d ) 1 + 3 ( x − a ) 1 ( x − c ) 1 ( x − d ) 1 + 5 ( x − a ) 1 ( x − b ) 1 ( x − d ) 1 + 7 ( x − a ) 1 ( x − b ) 1 ( x − d ) 1 } n o w i n b r a c k e t s i s a c u b i c e q u a t i o n − l e t t h e e x p r e s s i o n i n { } b r a c k e t s b e g ( x ) , f o r a < b < c < d − g ( a ) = n e g a t i v e g ( b ) = p o s i t i v e g ( c ) = n e g a t i v e g ( d ) = p o s i t i v e c l e a r l y g ( x ) h a s 3 r e a l a n d d i s i n c t r o o t s b y d r a w i n g a r o u g h g r a p h a n d a l s o i n 1 s t d e r i v a t i v e o f f ( x ) w e h a v e 3 m o r e s o l u t i o n s , t h e r e f o r e t o t a l n o . o f s o l u t i o n s a r e − 3 + 3 = 6
Draw the graph of f(x) and consider these points.
Between a and b, there's a minimum.
Then, at b, there are two roots of f'(x), but we shall count it only as one distinct root.
Then, between b and c, there's a maximum. So, that's another root.
Then, at c, you have 4 roots of the derivative that shall only be counted as one root.
Between c and d, you have one minimum.
Again, at d, you have 6 roots that shall only be counted as one distinct root of the derivative.
Thus you have 6 distinct roots.
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How do you know that you have found all the roots of f ′ ( x ) ?
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What do you mean? It seemed pretty clear to me from the graph and after differentiating once. If you count all the ones that I mentioned, but excluded because of multiplicities, you get 15 roots of f'(x).
Any polynomial in x that is of degree n has a derivative that has n-1 roots that may be real or complex.
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Yes, thats the right argument... it needs to be stated: You have found 15 roots already, counting multiplicities, so there can't be any more.
Being "pretty clear from the graph" is not such a convincing argument. It is conceivable that there might be additional points where f ′ ( x ) = 0 .
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By Rolle's Theorem, there are zeroes of f ′ ( x ) between a and b, between b and c, and between c and d.
Also, f ′ ( x ) has roots at b, c, and d with multiplicities 2, 4, and 6 respectively.
Counting roots with their multiplicities, we have found 15 roots of f ′ ( x ) , a polynomial of degree 15, so that there can be no more.
Thus f ′ ( x ) has 6 distinct roots.