Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.2 The Derivative as a Function - Exercises - Page 115: 53


$ p(x)=x^2+2x-3$.

Work Step by Step

We have $$ p(x)=x^2+ax+b, \quad p'(x)= 2x+a.$$ Since, $ p(1)=0$ and $ p'(1)=4$, we get the equations $$ a+b=-1, \quad a=2$$ Hence we have the solution $ a=2, b=-3$. Then, $ p(x)=x^2+2x-3$.
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