Question

What is the equation of the tangent line of `f(x)=xsqrt(x)` that is parallel to `g(x)=1+3x`?

Answer 1

`y=3x-4`

Explanation:

If we want a tangent line that is parallel to `g(x)=1+3x`, we know the tangent line will have the same slope as `1+3x`, which has a slope of `3`.

To find when the slope of the tangent line of `f(x)=xsqrtx` is equal to `3`, find the derivative of `f(x)`, since the derivative tells us the slope of the tangent line at any point.

To find the derivative of `f(x)`, first simplify `f(x)`.

`f(x)=xsqrtx=x^1(x^(1/2))=x^(3/2)`

Find `f'(x)` through the power rule.

`f'(x)=3/2x^(1/2)=(3sqrtx)/2`

So, we want to find when the tangent line has a slope of `3`, or when `f'(x)=3`.

`(3sqrtx)/2=3`

`sqrtx=2`

`x=4`

We know the tangent line of `f(x)` at `x=4` has slope `3`. To find the equation of the tangent line, find the point the line will intercept on `f(x)`.

`f(4)=4sqrt4=4(2)=8`

The point `(4,8)` and slope `3` can be related as a linear equation in point-slope form:

`y-8=3(x-4)`

Which can also be written as

`y=3x-4`