Eddie Clarke

2022-04-09

What is the equation of the normal line of $f\left(x\right)=\sqrt{{x}^{2}-x}$ at x=2?

titemomo8gjz

Beginner2022-04-10Added 10 answers

The normal line will be perpendicular to the tangent line when x=2.

We can determine what point on f(x) the normal line will intersect by finding that

$f\left(2\right)=\sqrt{2}$, so the point is $(2,\sqrt{2})$.

If we already have a point on the normal line, all we need to know is its slope. We can find of the tangent line when x=2 by finding f'(2). Since the normal line is perpendicular to the tangent line, its slope will be the opposite reciprocal of the tangent line's.

$f\left(x\right)={({x}^{2}-x)}^{\frac{1}{2}}$

Finding f'(x) will require use of the chain rule.

${f}^{\prime}\left(x\right)=\frac{1}{2}{({x}^{2}-x)}^{-\frac{1}{2}}\frac{d}{dx}[{x}^{2}-x]$

${f}^{\text{'}}\left(x\right)=\frac{1}{2}{({x}^{2}-x)}^{-\frac{1}{2}}(2x-1)$

$f}^{\prime}\left(x\right)=\frac{2x-1}{2\sqrt{{x}^{2}-x}$

Find the slope of the tangent line.

$f}^{\prime}\left(2\right)=\frac{2\left(2\right)-1}{2\sqrt{{2}^{2}-2}}=\frac{3}{2\sqrt{2}$

Take the opposite reciprocal to find that the slope of the normal line is $-\frac{2\sqrt{2}}{3}$.

Remember that it passes through the point $(2,\sqrt{2})$.

Write the equation in point-slope form:

$y-\sqrt{2}=-\frac{2\sqrt{2}}{3}(x-2)$

In slope-intercept form:

$y=-\frac{2\sqrt{2}}{3}x+\frac{7\sqrt{2}}{3}$

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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