What is the equation of the line that is normal to #f(x)= x/sqrt( 3x+2) # at # x=4 #?

Question

What is the equation of the line that is normal to #f(x)=  x/sqrt(  3x+2)    # at #  x=4  #?

Samantha Adkison · Accepted Answer

<img alt="enter image source here" src="https://tutor.hix.ai Do this: 
1)Find f'(x) 
This is the slope of the tangent line to the graph of f(x). Let's call it m_t. 
You want to find m_n, which is -1/(m_t). Recall if two lines are perpendicular, their respective slopes are negative reciprocals of one another. I've inserted a pretty TI screenshot, displaying the graph of f(x) with tangent line and normal line intersection. 
What is that point of intersection? 
2)Find the point (4,f(4)). 
3)Use the point-slope equation for a line 
y-y1=(m_n)(x-x1), where x1=4 and y1=f(4) and (m_n)=f'(4) 
That's your equation for the line. 
Your f'(x) should look like this....I don't want to do a step by step, cause that's your job... 
<mathjax>#(3x+4)/(2*(3x+2)^(3/2))#

Isabella Ackerman · Answer

undefined To find the equation of the line that is normal to the function f(x) = x/√(3x+2) at x=4, we need to determine the slope of the tangent line at x=4 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) with respect to x, which is given by:

f'(x) = (3x - 2)/(2√(3x+2))

Next, we substitute x=4 into the derivative to find the slope of the tangent line at x=4:

f'(4) = (3(4) - 2)/(2√(3(4)+2))
       = (12 - 2)/(2√(14))
       = 10/(2√(14))
       = 5/√(14)

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

m_normal = -1/(5/√(14))
         = -√(14)/5

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 (4, f(4)), so we substitute x=4 into the original function to find the corresponding y-coordinate:

f(4) = 4/√(3(4)+2)
     = 4/√(12+2)
     = 4/√14

Using the point-slope form, the equation of the line that is normal to f(x) at x=4 is:

y - (4/√14) = (-√14/5)(x - 4)