Question

For `f(x) =sqrt(x)^3`, what is the equation of the line tangent to the graph of `f` at the point where `x =125`?

Answer 1

`y = (15sqrt5)/2x-(625sqrt5)/2`

Explanation:

At `x=125`, `y = f(125) = (sqrt125)^3 = (5sqrt5)^3 = 125 * 5sqrt5 = 625 sqrt5`

`f(x) = x^(3/2)`, so `f'(x) = 3/2 x^(1/2)` and

at `x=125`, the slope of the tangent line is `m = f'(125) = (15sqrt5)/2`

The equation of the line through the point `(125, 625sqrt5)` with slope `m = (15sqrt5)/2` is

`y = (15sqrt5)/2x-(625sqrt5)/2`. (In slope-intercept form).

Answer 2

`y-125^(3/2)=15/2sqrt5(x-125)`

Explanation:

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

To find the equation of the tangent line, we need a point and a slope. The point would be at `(125,f(125))`, and the slope is `f'(125)`

Point: `(125,f(125))`
`(125,125^(3/2))`
`(125,1397.54)`

Slope:
`f'(x)=3/2x^(1/2)`
`f'(125)=3/2sqrt125`
`=15/2sqrt5`

Equation:
`y-125^(3/2)=15/2sqrt5(x-125)`