gG+ 3g + 2 = 0±²g³ = gG+ 3g + 2g = 0,±²0³ = 2 > 0, g = −1,±²−1³ = ²−1³G+ 3²−1³ + 2= −1 − 3 + 2 = −2 < 0´²g³ = ²g + 2³µ´¶²g³ = lim·→¸1ℎ¹²g + ℎ + 2³µ− ²g + 2³µº= lim·→¸1ℎ²gµ+ ℎµ+ 4 + 2ℎg+ 4g + 4ℎ − gµ− 4g − 4³= lim·→¸1ℎ²ℎµ+ 2ℎg + 4ℎ³= lim·→¸²ℎ + 2g + 4³ = 2g + 4Math 191, Quiz 4, Version A 1.Prove that the equation below has a solution The function ±²g³is continuous, so we can use Intermediate Value Theorem. We pick some values of gand find values of ±²g³. Function ±²g³is continuous on interval ¹−1, 0ºand changes sign between -1 and 0 so it has to be equal to zero somewhere within that interval, and the equation have a solution. 2.Use the definition of a derivative to find the derivative of: gG+ 2g + 1 = 0±²g³ = gG+ 2g + 1g = 0,±²0³ = 1 > 0,
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