written assignment 3

# written assignment 3 - c=-15-4b Then, by plugging in b for...

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JP Rodriguez Prof. Muralee Calc 1 Written Assignment A) To find the horizontal tangent, we need to find where y=0. We are given that f(x) = e x p(x) and p(x) = x 2 +x- 1. We then must plug p(x) into f(x) to get: e x (x 2 +bx+c)=0 Then we have to use the Chain Rule in order to get the derivatives of the two functions. The chain rule is (fg)’ = fg’+f’g, this gives us e x (x 2 +bx+c) + e x (2x+b)=0 Which simplifies to: 2e x (x 2 +(2+b)x+b+c)=0 We then divide out e x , leaving us with: x 2 +(2+b)x+b+c=0 And that is where the tangent is horizontal B) In this section, we are looking for the values of b and c where f(x) has horizontal tangents at x= -2, x=3. To do this, we first need to take our answer form the previous part, x 2 +(2+b)x+b+c=0. Then we plug in -2 and 3 for x. First, (-2) 2 +(2+b)(-2)+b+c=0 This will simplify to c=b Then, we plug in 3 (3) 2 +(2+b)(3)+b+c=0 Which will then simplify to

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Unformatted text preview: c=-15-4b Then, by plugging in b for c, we get an equation that we can solve for b= -15-4b 0= -15-5b 15= -5b-3=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) 2-4(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) 2-4(b+c))= -(2+b) - sqrt((2+b) 2-4(b+c)) Looking at this, it becomes evident that there can only be a horizontal tangent if (2+b) 2-4(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 .

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written assignment 3 - c=-15-4b Then, by plugging in b for...

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