A calculus problem by vishnu c

Calculus Level 5

Let f ( x ) = ( x a ) ( x b ) 3 ( x c ) 5 ( x d ) 7 f(x) = (x-a)(x-b)^3 (x-c)^5 (x-d)^7 , where a , b , c , d a,b,c,d real numbers with a < b < c < d a<b<c<d .

Consider the derivative of f ( x ) f(x) , f ( x ) f'(x) . Find the number of distinct real roots of f ( x ) = 0 f'(x) = 0 .


The answer is 6.

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3 solutions

Otto Bretscher
May 10, 2015

By Rolle's Theorem, there are zeroes of f ( x ) f'(x) between a and b, between b and c, and between c and d.

Also, f ( x ) 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 ) f'(x) , a polynomial of degree 15, so that there can be no more.

Thus f ( x ) f'(x) has 6 \boxed{6} distinct roots.

Righved K
Dec 8, 2015

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 f\left( x \right) =\left( { x-a } \right) ^{ 1 }\left( { x-b } \right) ^{ 3 }\left( { x-c } \right) ^{ 5 }\left( { x-d } \right) ^{ 7 }\\ f^{ / }\left( x \right) =\left( { x-b } \right) ^{ 2 }\left( { x-c } \right) ^{ 4 }\left( { x-d } \right) ^{ 6 }\{ \left( { x-b } \right) ^{ 1 }\left( { x-c } \right) ^{ 1 }\left( { x-d } \right) ^{ 1 }+3\left( { x-a } \right) ^{ 1 }\left( { x-c } \right) ^{ 1 }\left( { x-d } \right) ^{ 1 }+5\left( { x-a } \right) ^{ 1 }\left( { x-b } \right) ^{ 1 }\left( { x-d } \right) ^{ 1 }+7\left( { x-a } \right) ^{ 1 }\left( { x-b } \right) ^{ 1 }\left( { x-d } \right) ^{ 1 }\} \\ now\quad in\quad brackets\quad is\quad a\quad cubic\quad equation-\\ let\quad the\quad expression\quad in\quad \{ \} \quad brackets\quad be\quad g\left( x \right) ,\quad for\quad a<b<c<d-\\ g\left( a \right) =negative\\ g\left( b \right) =positive\\ g\left( c \right) =negative\\ g\left( d \right) =positive\\ \\ clearly\quad g(x)\quad has\quad 3\quad real\quad and\quad disinct\quad roots\quad by\quad drawing\quad a\quad rough\quad graph\quad and\quad also\quad in\quad 1st\quad derivative\quad of\quad f(x)\quad we\quad have\quad 3\quad more\quad solutions,\\ therefore\quad total\quad no.\quad of\quad solutions\quad are-\\ 3+3=6

Vishnu C
May 10, 2015

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 ) f'(x) ?

Otto Bretscher - 6 years, 1 month ago

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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.

vishnu c - 6 years, 1 month ago

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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 f'(x)=0 .

Otto Bretscher - 6 years, 1 month ago

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