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

Question

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

Christopher Allers · Accepted Answer

#"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)#

Isaiah Allen · Answer

undefined To find the slope of the line normal to the tangent line of f(x) = x^2-3sqrt(x) at x=3, we need to find the derivative of the function at x=3 and then take the negative reciprocal of that value.

First, let's find the derivative of f(x) = x^2-3sqrt(x).

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

Now, let's substitute x=3 into the derivative to find the slope of the tangent line at x=3.

f'(3) = 2(3) - (3/2)(1/sqrt(3))
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