What is the equation of the tangent line of #f(x)=sqrt((x-1)^3e^(2x) # at #x=2#?

Question

What is the equation of the tangent line of #f(x)=sqrt((x-1)^3e^(2x)   # at #x=2#?

Elijah Abram · Accepted Answer

#g (x)=5/2 e^2x -4e^2#

To solve this problem as presented, we must do the following:

(1)

#f (2)= sqrt ((2-1)^3e^(2*2)) = sqrt (e^4) = e^2#

(2)

#f (x)= sqrt ((x-1)^3e^(2x)) = e^x(x-1)^(3/2)#

Recall that the product rule states for

#a (x)=b (x)c (x) => a'=b'c+bc'#

Additionally, the chain rule states that given

#k (x)=m (n (x)), (dk)/dx = (dn)/dx*(dm)/(dn) #.

Here, we have the following:

#a (x) = f (x) = b (x)c (x)=e^x (x-1)^(3/2)#

#=> b (x) = e^x -> b'(x) = e^x#

#=> c (x) = k (x) = m (n (x)) = (x-1)^(3/2)#

where we take

#n (x)=(x-1) ->(dn)/dx = 1#

#m (n) = n^(3/2) -> (dm)/(dn) = 3/2 n^(1/2)#

#c'(x) = (dc)/dx = (dk)/dx = 3/2 (x-1)^(1/2)#

#f'(x) = e^x (x-1)^(1/2) + 3/2 (x-1)^(1/2) e^x#

#f'(2) = e^2 +3/2 e^2 = 5/2 e^2#

(3) We now find #g(x)#, which we define as being the tangent line at #x=2#. We use our slope from (2), our #f (2)# from (1), and our #x# from the problem to get...

#g(x) - e^2 = 5/2 e^2 (x-2) = 5/2 e^2x-4e^2 -> g (x)=5/2 e^2x -4e^2#

Isaiah Allen · Answer

The equation of the tangent line is

#y=5/2e^2x - 4e^2# The equation of a line in the form #y=mx+b# contains a slope, #m# and a #y#-intercept #b#. We can obtain the slope from the equation by taking the derivative and substituting the #x# value of #2#. We can then solve for the #y#-intercept using the point #(2,f(2))# Let's begin by getting the y-value of the point we want the line to go through: #f(2) = sqrt((2-1)^3*e^(2*2))# #f(2) = sqrt((1)^3*e^(4)) = e^2# Next we need to take the derivative of #f(x)#, but first I'd like to distribute the square root over the factors inside it, just so I don't have to use the chain rule an extra time: #f(x) = (x-1)^(3/2)*e^x# #f'(x) = d/(dx)[(x-1)^(3/2)*e^x] # Then we use the chain rule: #f'(x) = [d/(dx) (x-1)^(3/2)] * e^x + (x-1)^(3/2) * [d/(dx)e^x]# #f'(x) = 3/2 (x-1)^(1/2) * e^x + (x-1)^(3/2)*e^x# solving for the slope at the point of interest: #f'(2) = 3/2 (2-1)^(1/2) * e^2 + (2-1)^(3/2)*e^2# #f'(2) = 3/2 e^2 + e^2 = 5/2e^2# Now in our equation for the line at the point #(2,e^2)#: #y = 5/2e^2x +b# #e^2 = 5e^2 + b# # => b=-4e^2# finally the equation of the tangent line is #y=5/2e^2x - 4e^2# which we can graph to check our solution: graph{(5(e^2)/2x-4e^2-y)(sqrt((x-1)^3e^(2x))-y)=0 [1 3 -20 20]}

Elijah Abram · Answer

undefined To find the equation of the tangent line of f(x) = sqrt((x-1)^3e^(2x)) at x = 2, we need to find the slope of the tangent line and a point on the line.

First, we find the derivative of f(x) using the chain rule and product rule. The derivative is given by:

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

Next, we substitute x = 2 into f'(x) to find the slope of the tangent line:

f'(2) = (3(2-1)^2e^(2*2) + 2(2-1)^3e^(2*2))/(2sqrt((2-1)^3e^(2*2)))

Simplifying this expression gives us the slope of the tangent line.

Finally, we use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, to find the equation of the tangent line. We substitute the values of x1, y1, and m into the equation and simplify to obtain the final equation of the tangent line.