Calculus 10th Edition

Published by Brooks Cole

Chapter 2 - Differentiation - 2.4 Exercises: 84

Answer

At one point: $(1, 1).$

Work Step by Step

Using the quotient rule: $f’(x)=(\frac{u(x)}{v(x)})'=\frac{u'(x)v(x)-v'(x)u(x)}{(v(x))^2}$ $u(x)=x; u'(x)=1$ $v(x)=\sqrt{2x-1}; v'(x)=\dfrac{1}{\sqrt{2x-1}}$ $f'(x)=\dfrac{\sqrt{2x-1}-\dfrac{x}{2x-1}}{2x-1}$ $=\dfrac{x-1}{\sqrt{(2x-1)^3}}$ $f'(x)=0 \rightarrow \dfrac{x-1}{\sqrt{(2x-1)^3}}=0 \rightarrow x-1=0 \rightarrow x=1$ $f(1)=\dfrac{1}{\sqrt{2(1)-1}}=1$ Hence the point at which there is a horizontal tangent is $(1, 1).$

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