What is the equation of the line normal to f(x)=4x3−1 at x=0?

Given that f(x)=4x3−1 and x0=0y0=f(0)=−4The slope of the normal line at x=x0 is the negative reciprocal of the derivative of the function,evaluated at x=x0:M(x0)=−1f′(x0)Find the derivative: f′(x)=(4x3−1)′=−12x2(x3−1)2Hence, M(x0)=−1f′(x0)=(x03−1)212x02Next, find the slope at the given point.As can be seen, the slope is infinite.Since the slope is not finite, the equation of the normal line is x=x0=0.

Solved: "What is the equation of the line normal..."

Tristian Neal

Answered question

2022-04-07

What is the equation of the line normal to

f (x) = \frac{4}{x^{3} - 1}

at x=0?

Answer & Explanation

withthedevilwry5

Beginner2022-04-08Added 6 answers

f (x) = \frac{4}{x^{3} - 1} = 4 {(x^{3} - 1)}^{- 1}

f^{'} (x) = - 4 {(x^{3} - 1)}^{- 2} (3 x^{2})

f^{'} (0) = - 4 {(0^{3} - 1)}^{- 2} (3 {(0)}^{2}) = 0

f'(0)=0 means the tangent line is horizontal
the normal line then must be vertical (line is in the form x=a)
since the point (0, f(0)) lies on the vertical line x=0, that is the normal line

Kendall Wilkinson

Beginner2022-04-09Added 17 answers

Given that

f (x) = \frac{4}{x^{3} - 1}

and

x_{0} = 0

y_{0} = f (0) = - 4

The slope of the normal line at

x = x_{0}

is the negative reciprocal of the derivative of the function,
evaluated at

x = x_{0} : M (x_{0}) = - \frac{1}{f^{'} (x_{0})}

Find the derivative:

f^{'} (x) = {(\frac{4}{x^{3} - 1})}^{'} = - \frac{12 x^{2}}{{(x^{3} - 1)}^{2}}

Hence,

M (x_{0}) = - \frac{1}{f^{'} (x_{0})} = \frac{{(x_{0}^{3} - 1)}^{2}}{12 x_{0}^{2}}

Next, find the slope at the given point.
As can be seen, the slope is infinite.
Since the slope is not finite, the equation of the normal line is

x = x_{0} = 0

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