What is the slope of the line normal to the tangent line of $f(x) = x^2-3sqrtx$ at $x= 3$ ?

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Answer 1 · 2018-05-01T10:00:17

$"Slope of Normal"=-(24+2*sqrt(3))/(141)$

Rewrite the square root: $sqrt(x)=x^(1/2)$

Apply the power rule twice to find the first derivative of $f(x)$ .

$f'(x)=2x-3*1/2*x^(-1/2)$
$f'(x)=2x-3/(2sqrt(x))$

Evaluating $f'(x)$ at $x=3$ gives the slope of the tangent

$f'(3)=2*3-3/(2sqrt(3))=1/2*(12-sqrt(3))$

The gradient product of two slant lines perpendicular to each other should equal to $-1$ , that is: they form a pair of negative reciprocals.

Hence, the slope of the normal would equal to the opposite of the slope of the tangent inversed

$"Slope of Normal"=-1/("Slope of Tangent")$

$" "color(white)(l)=-1/(1/2*(12-sqrt(3)))$

Simplifying

$-1/(1/2*(12-sqrt(3)))$
$=2*1/(sqrt(3)-12)*color(grey)((sqrt(3)+12)/(sqrt(3)+12))$
$=-(24+2*sqrt(3))/(141)$

What is the slope of the line normal to the tangent line of f(x) = x^2-3sqrtx f(x) = x^2-3sqrtx at x= 3 x= 3 ?