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Brendan Caseria MATH 2413: Dallas College Due: 10/22/2023 O o0 Esginatay Homework 8: 3.10, 4.1, 4.2 Question 1. Suppose a particle starts at the origin and moves along the curve y = 3\/Z in the first quadrant (i.e. both z and y are increasing). Suppose that when z = 16, the particle is moving away from the origin at a rate of 20 units per second. a) What is & at this moment in time? b) What is 9 at this moment in time? Y |3 Xy (3) =y 3Joe) =12 o Tq)k:kd'" C"O/\Z) dr qd;& 24 =T == d_g_ |L1¥\2,1<L$z e it e 256yinh -_—qz 2('5)[20\'\4(,20) _Z(?O-) Cuo+180 =402 \{00:«0 1 Y= 20 4920) :4@ ye r -d4 ZOtS - —d—¥ Scanned with CamScanner

2 Question 2. Our goal is to prove that f(z) = z + e* has exactly one root (i.e. f(z) =0 for exactly one value of z). Note: This is VERY similar to Ezample 1 from Section 4.2 in the book. I might suggest you read that before proceeding, if you find yourself struggling. That e:n:ample has a condensed solution; we're going to flesh it out a bit more. a) First, we really need this function to be both continuous and differentiable. Explain in words how we know this is true. fehy L continees evcn:st,mem,'rhctc?ccc-nmc\\(:na % 4 diffcrentiavic . , . . b) Now, let's prove that f(z) has at least one root. What theorem do we need for that? Hint: This is a review problem! We're looking all the way back at section 2.5. larecmidiare Valye Theorewn / IVT ¢) Use the theorem you stated in part a) to prove that f(z) has at least one root. So b'é IVT , neee exisy C =\ berwetr | and =lre' Sydn vner Bee) = O Scanned with CamScanner

3 d) Now that we've proven there is at least one root, we will now prove there is not at least two roots. We'll do this using contradiction. Let's (incorrectly) assume that there are in fact two different roots, and use that to blow up math. The logic here is that if we assume something, and end up breaking math using that assumption, then our assumption should be wrong and we can correctly conclude its opposite. Let's call those two roots a and b. Mean Value Theorem then states that the derivative of f(z) should equal L8202 somewhere. What is £8=1(2) numerically equal to in this case? Assume aLb soon vhet fea) =0 and $ce) =0 - €0y s conv, over Lo, 0T} ona oF. over (ob). So by MVT dhee exsts ¢ berwcen 0 ond b, Som thee fCy = FeB) -y v T(C) boo - © ¢) Calculate f/(z) and find its range. Does your answer from part d) fall in the range? In other words, even though MVT said this must happen, is it actually true that f'(z) can attain the value found in part d)? 'f'(A) = red and $TIZ2 | ¥0 =) e onsWeC S0 pacy d does ner o\ ia ™e fage betovse fR)X | ond O s Vess. Ffeardz xyer B 4 £00)9 1 1-;':' f) If you did parts d) and e) correctly, then congratulations, you created a contradiction! What can we now conclude? Trece orec NOr 2 (oors and ynere s CAacrly 1 root. Scanned with CamScanner