Question

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

Answer 1

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

Explanation:

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

1. Find a point in the line (i.e. find `f(2)`).
2. Find `f'(x)` (by use of product and chain rules)
3. Recalling that (2) gives us the slope of the tangent line, use point slope form with `m=f'(2), y_1 = f (2), x_1 =2`

(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`

Answer 2

The equation of the tangent line is

`y=5/2e^2x - 4e^2`

Explanation:

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]}