What is the equation of the normal line of f(x)=x3−x+4 at x=2?

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titemomo8gjz · Accepted Answer

We begin by finding the derivative of the function f(x) using the chain rule

f^{'} (x) = \frac{1}{2} \cdot \frac{1}{\sqrt{x^{3} - x + 4}} \cdot (3 x^{2} - 1)

The slope of the tangent line at x=2 is then

f^{'} (2) = \frac{1}{2} \cdot \frac{1}{\sqrt{8 - 2 + 4}} \cdot (3 \cdot 4 - 1) = \frac{11}{2 \sqrt{10}}

The slope of the normal line is therefore the negative reciprocal of this slope:

m_{p} = - {\frac{1}{f}}^{'} (2) = - \frac{2 \sqrt{10}}{11}

To get the equation of the line, we need to find a point to substitute into the equation to get the y-intercept - we need the value of the function at x=2

f (2) = \sqrt{8 - 2 + 4} = \sqrt{10}

Substituting this into the equation of the line we obtain

\sqrt{10} = - (2 \frac{\sqrt{10}}{11}) \cdot 2 + b

Solving for b we get

b = \frac{15 \sqrt{10}}{11}

Therefore the equation of the normal line to our function at x=2 is

y = - \frac{2 \sqrt{10}}{11} x + \frac{15 \sqrt{10}}{11}

What is the equation of the normal line

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