What is the equation of the line normal to f(x)=2x3−x at x=1?

Let y=2x3−xy=(2x3−x)12Applying the chain rule ⇒y′=12(2x3−x)−12×(6x2−1)y′=(6x2−1)22x3−xThis equals the gradient m.So when x=1 ⇒m=(6−1)22−1=52The tangent line is of the form:y=mx+cAt x=1,y=2−1=1∴1=52×1+c∴c=−52So the equation of the tangent is:y=52x−52

Key to "What is the equation of the line normal..."

High School
Calculus and Analysis
Calculus 1
Derivatives

Charlotte Holden

Answered question

2022-04-08

What is the equation of the line normal to

f (x) = \sqrt{2 x^{3} - x}

at x=1?

Answer & Explanation

glanzerjbdo

Beginner2022-04-09Added 13 answers

Let

y = \sqrt{2 x^{3} - x}

y = {(2 x^{3} - x)}^{\frac{1}{2}}

Applying the chain rule

\Rightarrow

y^{'} = \frac{1}{2} {(2 x^{3} - x)}^{- \frac{1}{2}} \times (6 x^{2} - 1)

y^{'} = \frac{(6 x^{2} - 1)}{2 \sqrt{2 x^{3} - x}}

This equals the gradient m.
So when x=1

\Rightarrow

m = \frac{(6 - 1)}{2 \sqrt{2 - 1}} = \frac{5}{2}

The tangent line is of the form:
y=mx+c
At x=1,

y = \sqrt{2 - 1} = 1

∴ 1 = \frac{5}{2} \times 1 + c

∴ c = - \frac{5}{2}

So the equation of the tangent is:

y = \frac{5}{2} x - \frac{5}{2}

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