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Unformatted text preview: c=154b Then, by plugging in b for c, we get an equation that we can solve for b= 154b 0= 155b 15= 5b3=b And since b=c, the value for both variables that will give us a horizontal tangent is b= 3 c= 3 C) Now we must find if there is a quadratic f(x)=e x p(x) that has only one horizontal tangent, and if it does, what restrictions would this put on b and c Once again we need to take the answer from the first equation x 2 +(2+b)x+b+c=0. Then we use the quadratic formula. ((2+b) +/ sqrt((2+b) 24(b+c)))/2 Then we can look at just the numerator and set both the plus and the minus equal to each other(2+b) + sqrt((2+b) 24(b+c))= (2+b)  sqrt((2+b) 24(b+c)) Looking at this, it becomes evident that there can only be a horizontal tangent if (2+b) 24(b+c) = 0...
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This note was uploaded on 04/13/2008 for the course MATH 021 taught by Professor Muralee during the Spring '08 term at Lehigh University .
 Spring '08
 Muralee

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