## Calculus 8th Edition

Published by Cengage

# Chapter 10 - Parametric Equations and Polar Coordinates - Review - Exercises - Page 730: 27

#### Answer

Point corresponding to $t=-\frac{1}{2}$ is $$(\frac{11}{8},\frac{3}{4})$$ or, $$(1.35, 0.75)$$ as shown in the attached graph.

#### Work Step by Step

As we are given that $$f(t)=t^2+t+1$$ It will be minimum for $$f'(t)=2t+1$$ $f'(t)=0$ when $t=-\frac{1}{2}$ and $$f''(t)=2$$ As per second derivative test , $f(t)$ is minimum when $t=-\frac{1}{2}$ . Hence, point corresponding to $t=-\frac{1}{2}$ is: $$(\frac{11}{8},\frac{3}{4})$$ or, $$(1.35, 0.75)$$.

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