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

Question

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

Elijah Abram · Accepted Answer

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=- sqrt(7)/9(x-41)# 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(x)=sqrt(x^3-3x+2)=(x^3-3x+2)^(1/2)# #f'(x)=1/2(x^3-3x+2)^(-1/2)(3x^2-3)# When x=5, the gradient of the tangent line is: #f'(5)=1/2(125-15+2)^(-1/2)(75-3)# #f'(5)=1/2(112)^(-1/2)(72)=36/(4sqrt(7))=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= #- sqrt(7)/9# When x=5, y at the point where f(x) and the line meet will be f(5): #y=f(5)=sqrt((5)^3-3(5)+2)=sqrt(125-15+2)=4sqrt(7)# Now we can find the y-intercept, c. #c= y - mx = 4sqrt(7) + (sqrt(7)(5))/9=(41sqrt(7)) /9# We can now finally write the normal line equation: #y = - sqrt(7)/9 x + (41sqrt(7)) /9# #y=- sqrt(7)/9(x-41)#

Isaiah Allen · Answer

undefined To find the equation of the line that is normal to the function f(x) = sqrt(x^3 - 3x + 2) at x = 5, we need to determine the slope of the tangent line at x = 5 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can take the derivative of the function f(x) with respect to x and evaluate it at x = 5.

The derivative of f(x) = sqrt(x^3 - 3x + 2) is given by:

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

Evaluating f'(x) at x = 5, we get:

f'(5) = (3 * 5^2 - 3) / (2 * sqrt(5^3 - 3 * 5 + 2))

Simplifying further:

f'(5) = (3 * 25 - 3) / (2 * sqrt(125 - 15 + 2))
      = (72) / (2 * sqrt(112))
      = 36 / sqrt(112)

The slope of the tangent line at x = 5 is 36 / sqrt(112).

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line:

Slope of normal line = -1 / (36 / sqrt(112))
                    = -sqrt(112) / 36

Now that we have the slope of the normal line, we can use the point-slope form of a line to find the equation. We know that the line passes through the point (5, f(5)).

Substituting the values into the point-slope form:

y - f(5) = (-sqrt(112) / 36) * (x - 5)

Simplifying further:

y - f(5) = (-sqrt(112) / 36) * x + (5 * sqrt(112) / 36)

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