Calculus - Inadequate Information #2
There is a cubic polynomial f ( x ) . Some real constants a , b and c satisfy
{ f ( a ) − b } { a f ′ ( c ) − b + f ( c ) − c f ′ ( c ) } < 0 ; f ′ ′ ( c ) = 0 .
at how many points do the graphs of y = ( a − x ) f ′ ( x ) and y = − f ( x ) + b meet?
This problem is a part of <Calculus - Inadequate Information> series .
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1 solution
Wow! Such an abstraction from basics! Nice problem +1
Great problem! Wasted me like a week ;)
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Heh x'D It's not wasting time, it's having fun! :')
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Nah... Do agree with what you said, but what I meant was: "Hell... Could have done this a week earlier!" XD
@H.M. 유 I did it like this :
Define a function T ( x ) as :
T ( x ) = ( a − x ) f ′ ( x ) + f ( x ) − b
T ′ ( x ) = ( a − x ) f ′ ′ ( x )
T ( a ) = f ( a ) − b and T ( c ) = ( a − c ) f ′ ( c ) + f ( c ) − b
According to the given information :
T ( a ) T ( c ) < 0
Hence , there exist atleast one solution in ( a , c ) or ( c , a )
Moreover :
T ′ ( c ) = T ′ ( a ) = 0 ⟹ T ′ ( x ) = p ( x − a ) ( x − c )
⟹ T ′ ′ ( a ) T ′ ′ ( c ) < 0
Hence , two more solutions exist in ( − ∞ , a ) ∪ ( c , ∞ ) or ( − ∞ , c ) ∪ ( a , ∞ )
Am I correct with the last two conditions ?
How to draw 3 tangents?
{ f ( a ) − b } { a f ′ ( c ) − b + f ( c ) − c f ′ ( c ) } can be written as:
{ b − f ( a ) } { b − ( ( a − c ) f ′ ( c ) + f ( c ) ) } < 0 , which is equivalent to:
⎩ ⎪ ⎨ ⎪ ⎧ f ( a ) < b < ( a − c ) f ′ ( c ) + f ( c ) ( f ( a ) < ( a − c ) f ′ ( c ) + f ( c ) ) ( a − c ) f ′ ( c ) + f ( c ) < b < f ( a ) ( f ( a ) > ( a − c ) f ′ ( c ) + f ( c ) )
Now let's figure out what in the hell this ( a − c ) f ′ ( c ) + f ( c ) is.
Since f ( x ) is a cubic polynomial, f ′ ′ ( x ) is linear and therefore has only one root, which is c .
Therefore the inflection point of f ( x ) is ( c , f ( c ) ) , and the inflection tangent line (a line that is tangent to the function at its inflection point) is y = ( x − c ) f ′ ( c ) + f ( c ) .
Then let's draw this on the coordinates system.
The pink area is where the point ( a , b ) can exist - you can imagine where the pink area would be if the leading coefficient of f ( x ) were negative.
We can see that any point in the pink area can draw exactly three tangent lines to the function.
Let's see the problem again.
What the problem is asking is the same as the number of real roots of b = ( a − t ) f ′ ( t ) + f ( t ) .
Note that y = ( x − t ) f ′ ( t ) + f ( t ) denotes the tangent line at ( t , f ( t ) ) .
Therefore the number of real roots of b = ( a − t ) f ′ ( t ) + f ( t ) is precisely the number of tangent lines that can be drawn from point ( a , b ) .
As we've already figured out, the answer is 3 .