What is the equation of the line normal to  f(x)=-xln(2x^3-x) at  x=1?

Dec 25, 2016

$y = \frac{1}{5} x - \frac{1}{5}$

Explanation:

The gradient of the tangent to a curve at any particular point is give by the derivative of the curve at that point. The normal is perpendicular to the tangent and so the product of their gradients is $- 1$

so If $f \left(x\right) = - x \ln \left(2 {x}^{3} - x\right)$ then differentiating wrt $x$ (using the product rule and chain rule) gives us:

$\setminus \setminus \setminus \setminus \setminus f ' \left(x\right) = \left(- x\right) \left(\frac{1}{2 {x}^{3} - x} \cdot \left(6 {x}^{2} - 1\right)\right) + \left(- 1\right) \left(\ln \left(2 {x}^{3} - x\right)\right)$
$\therefore f ' \left(x\right) = - \frac{6 {x}^{2} - 1}{2 {x}^{2} - 1} - \ln \left(2 {x}^{3} - x\right)$

When $x = 1 \implies f \left(1\right) = - 1 \ln \left(2 - 1\right) = 0$ (so $\left(1 , 0\right)$ lies on the curve)
and $f ' \left(1\right) = - \frac{6 - 1}{2 - 1} - \ln \left(2 - 1\right) = - 5$

So the tangent passes through $\left(1 , 0\right)$ and has gradient $- 5$, hence the normal has gradient $\frac{1}{5}$ so using the point/slope form $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the equation we seek is;

$y - 0 = \frac{1}{5} \left(x - 1\right)$
$\setminus \setminus \therefore y = \frac{1}{5} x - \frac{1}{5}$

We can confirm this solution is correct graphically: