{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

quiz 4 solutions

# quiz 4 solutions - Math 191 Quiz 4 Version A Math 191 Quiz...

This preview shows page 1. Sign up to view the full content.

g G + 3g + 2 = 0 ±²g³ = g G + 3g + 2 g = 0, ±²0³ = 2 > 0, g = −1, ±²−1³ = ²−1³ G + 3²−1³ + 2 = −1 − 3 + 2 = −2 < 0 ´²g³ = ²g + 2³ µ ´ ²g³ = lim ·→¸ 1 ¹²g + ℎ + 2³ µ − ²g + 2³ µ º = lim ·→¸ 1 ²g µ + ℎ µ + 4 + 2ℎg + 4g + 4ℎ − g µ − 4g − 4³ = lim ·→¸ 1 ²ℎ µ + 2ℎg + 4ℎ³ = lim ·→¸ ²ℎ + 2g + 4³ = 2g + 4 Math 191, Quiz 4, Version A 1. Prove that the equation below has a solution The function ±²g³ is continuous, so we can use Intermediate Value Theorem. We pick some values of g and find values of ±²g³ . Function ±²g³ is continuous on interval ¹−1, 0º and 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: g G + 2g + 1 = 0 ±²g³ = g G + 2g + 1 g = 0, ±²0³ = 1 > 0,
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online