Question

What is the equation of the line normal to `f(x)=xsqrtx-e^(sqrtx)` at `x=1`?

Answer 1

`y = -2/(3-e) (x - 1) + 1 - e`

Explanation:

Let's first find the y-value that corresponds with x = 1 on the function `f(x) = xsqrt(x) - e^(sqrt(x))`:

`f(1) = 1sqrt(1) - e^(sqrt(1))`
`f(1) = 1 - e`

Now, find the derivative of the given function to find an equation for the slope of the tangent line:

`f(x) = xsqrt(x) - e^(sqrt(x))`
`f(x) = x^(3/2) - e^(x^(1/2))` (rewriting f(x) to make it easier to differentiate)

`fprime(x) = 3/2x^(1/2) - e^(x^(1/2))*1/2x^(-1/2)`
`fprime(x) = (3sqrt(x))/2 - (e^sqrt(x))/(2sqrt(x))`

Substitute x = 1 to find the slope of the tangent line at that point:

`fprime(1) = slope = (3sqrt(1))/2 - (e^sqrt(1))/(2sqrt(1))`
`slope = (3-e)/2`

The slope of the normal line at a point is the negative reciprocal of the slope of the tangent line at that point. In this case, the slope of the normal line is `-2/(3-e)`.

We can now write the equation of the normal line:
`y - yo = m(x - xo)`
`y - 1 + e = -2/(3-e) (x - 1)`
`y = -2/(3-e) (x - 1) + 1 - e`