# What is the equation of the line that is normal to f(x)= sqrt( x^3-3x+2)  at  x=5 ?

Feb 13, 2016

The straight line equation is: y=mx+x. The gradient m is gained from differentiating f(x), then m when x=5 is f'(x=5). The value of y is gained from f(x=5). Now you have values for y,x and m, so can find c. Take the negative reciprocal of the tangent gradient to get the normal gradient, and the normal line equation is:
$y = - \frac{\sqrt{7}}{9} \left(x - 41\right)$

#### Explanation:

The straight line equation is: y=mx+x. The gradient m is gained from differentiating f(x). Thus m when x=5 is f'(x=5). Use the chain rule to perform this differentiation:
$f \left(x\right) = \sqrt{{x}^{3} - 3 x + 2} = {\left({x}^{3} - 3 x + 2\right)}^{\frac{1}{2}}$
$f ' \left(x\right) = \frac{1}{2} {\left({x}^{3} - 3 x + 2\right)}^{- \frac{1}{2}} \left(3 {x}^{2} - 3\right)$

When x=5, the gradient of the tangent line is:
$f ' \left(5\right) = \frac{1}{2} {\left(125 - 15 + 2\right)}^{- \frac{1}{2}} \left(75 - 3\right)$
$f ' \left(5\right) = \frac{1}{2} {\left(112\right)}^{- \frac{1}{2}} \left(72\right) = \frac{36}{4 \sqrt{7}} = \frac{9}{\sqrt{7}}$ = m at x=5.

Now to obtain the normal line gradient, we take the reciprocal of the tangent gradient:
For the normal, m= $- \frac{\sqrt{7}}{9}$

When x=5, y at the point where f(x) and the line meet will be f(5):
$y = f \left(5\right) = \sqrt{{\left(5\right)}^{3} - 3 \left(5\right) + 2} = \sqrt{125 - 15 + 2} = 4 \sqrt{7}$

Now we can find the y-intercept, c.

$c = y - m x = 4 \sqrt{7} + \frac{\sqrt{7} \left(5\right)}{9} = \frac{41 \sqrt{7}}{9}$

We can now finally write the normal line equation:
$y = - \frac{\sqrt{7}}{9} x + \frac{41 \sqrt{7}}{9}$
$y = - \frac{\sqrt{7}}{9} \left(x - 41\right)$