Calculus (3rd Edition)

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

Chapter 1 - Precalculus Review - Chapter Review Exercises - Page 37: 31


$f(x)$ has the minimum value $f(x)= 49/10$

Work Step by Step

We are given the points $(2,1)$ and $(x,3x+2)$. Recall the distance formula between two points: $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ We are told that the function is the square of the distance, so we square the formula above and plug in the given points to obtain: \begin{align*} f(x)&=(x-2)^{2}+(3 x+2-1)^{2}\\ &=x^{2}-4 x+4+9 x^{2}+6 x+1\\ &=10 x^{2}+2 x+5\\ &=10\left(x^{2}+\frac{1}{5} x+\frac{1}{100}\right)+5-\frac{1}{10}\\ &=10\left(x+\frac{1}{10}\right)^{2}+\frac{49}{10} \end{align*} Compare with the quadratic function $$f(x)=a(x-h)^{2}+k$$ Then the vertex is: $(h,k)= ( 1/10, 49/10)$ Since $a>0 $, then $f(x)$ has a minimum value of $f(x)= 49/10$
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