Derivatives: prove there is no horizontal tangent line to graph
5,327
Differentiate implicitly:
$$2x-3y-3xy'+2yy'=0\implies(2y-3x)y'=3y-2x\implies y'=\frac{3y-2x}{2y-3x}$$
Thus, the only point where the derivative is defined and equals zero is when $\;3y-2x=0\;,\;\;(x,y)\in\text{ Dom}\,(f)\;$ , but then
$$y=\frac23x\implies f\left(x,\,\frac23x\right)=x^2-2x^2+\frac49x^2=1\iff-\frac59x^2=1$$
and this last equality is impossible (in the real number, of course), thus $\;y'\neq0\;$ for all points where it is defined.
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Author by
Bob Umadbro
Updated on August 16, 2022Comments
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Bob Umadbro about 1 year
Equation: $-3xy + x^2 + y^2 = 1$
Demonstrate that there is no horizontal tangent using inverse variation.
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user247327 about 7 yearsYou mean, of course, that "$y'\ne 0$ for all points where it is defined".
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DonAntonio about 7 years@user247327 Yes, of course. Thanks.
