Calculus1
                        12 Online
    				
    				
                        
                        OpenStudy (emilyjones284):
                        Rolle's Theorem
f(x)=(x-2)(x-3)(x-4) on [2,4]
                        
                     
                    
                    
                    
                    
                    
    
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OpenStudy (perl):
did you satisfy the hypotheses of rolle's theorem, first?
 
OpenStudy (perl):
it must be differentiable on (2,4) and continuous on [2,4] , and it is since it is a polynomial 
 
OpenStudy (emilyjones284):
thats what I don't know how to do
 
OpenStudy (emilyjones284):
Oh okay I wasn't really sure that it was a polynomial
 
OpenStudy (perl):
its a polynomial in factored form (linear factors) 
 
    
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OpenStudy (emilyjones284):
so then plug in 2 and 4
 
OpenStudy (emilyjones284):
and then take the derivative to find c?
 
OpenStudy (perl):
first plug in f(2) and f(4)
 
OpenStudy (perl):
rolles theorem says we must have f(a) = f(b) 
 
OpenStudy (emilyjones284):
okay well I got -2 for f(2) and 0 for f(4) so rolle's theorem does not apply right?
 
    
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OpenStudy (perl):
should be zero for f(2) 
 
OpenStudy (perl):
if you plug in 2 , you have (2-2) * whatever = 0 
 
OpenStudy (emilyjones284):
oops i see what I did wrong
 
OpenStudy (emilyjones284):
so then i just take the derivative and find c right?
 
OpenStudy (perl):
so now since we have f(2) = f(4) , then there exists a value c inside (2,4) such that f ' (c) = 0 
 
    
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OpenStudy (perl):
correct, take derivative
 
OpenStudy (emilyjones284):
the derivative is just 1 right?
 
OpenStudy (perl):
the derivative should not be 1
 
OpenStudy (emilyjones284):
but the derivative for each is 1 isn't it?
 
OpenStudy (perl):
yes but youre using product rule incorrectly 
 
    
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OpenStudy (emilyjones284):
so then it's just (1)(1)(1)
 
OpenStudy (emilyjones284):
what am I doing wrong?
 
OpenStudy (perl):
well 
 
OpenStudy (perl):
you want to use product rule? 
 
OpenStudy (perl):
here is the product rule
( u * v ) ' = u'  * v + u * v ' 
 
    
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OpenStudy (emilyjones284):
yes I think
 
OpenStudy (perl):
we have three factors here though 
 
OpenStudy (emilyjones284):
ohhh I thought i couldve just taken the derivative of each
 
OpenStudy (perl):
( u * v * w ) ' = u ' * v  * w + u * v ' * w  + u * v * w ' 
 
OpenStudy (perl):
i dont think this is the easiest way to differentiate it , though 
 
    
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OpenStudy (emilyjones284):
ohh there's an easier way?
 
OpenStudy (perl):
yes we can expand the original f(x) 
 
OpenStudy (anonymous):
product rule?!
 
OpenStudy (perl):
technically you could use the product rule :) 
 
OpenStudy (perl):
but will be more work 
 
    
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OpenStudy (emilyjones284):
so what should I do?
 
OpenStudy (perl):
f(x)=(x-2)(x-3)(x-4)  
= ( x^2 -2x -3x + 6) (x-4)
 
OpenStudy (perl):
so first expand (x-2)(x-3) using foil 
 
OpenStudy (emilyjones284):
oh and then use the product rule from there?
 
OpenStudy (perl):
then multiply by x - 4 
 
    
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OpenStudy (perl):
(x-2)(x-3)(x-4) 
= ( x^2 -2x -3x + 6) (x-4)
= ( x^2 - 5x + 6) ( x - 4)
= x^3 - 5x^2 + 6x - 4x^2 +20x - 20 
 
OpenStudy (perl):
f(x) = x^3 -8x^2 + 26x - 20   
 
OpenStudy (perl):
ok? foil then foil again 
 
OpenStudy (emilyjones284):
and then I take the derivative?
 
OpenStudy (perl):
yes
 
    
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OpenStudy (perl):
now its easy to take the derivative 
 
OpenStudy (emilyjones284):
ohh okay thank you so much