Question

Given f'(x) = (x-4)(4-2x), find the x-coordinate for relative maximum on the graph of f(x)

Answer 1

The coordinate of the relative maximum is x=4.

Explanation:

Given that the derivative of the function
`f(x)` is
`f'(x) = (x-4)(4-2x)`, the maxima and minima or the critical points can be found where
`f'(x)=0,` that is:

`f'(x) = (x-4)(4-2x)=0.`

The solutions to this equation are
`x=4` and
`x=2.`

Now, if the second derivative
`f''(x)` for a function
`f(x)` is negative at a critical point, then the critical point is the relative maximum.

Therefore we want to see at which of the two critical points is
`f''(x)` negative. The second derivative is:

`\frac{\mathrm{d f'(x)}}{\mathrm{d}x}=f''(x)=-4(x-3).`

Now
`f''(4)` is
`-4`, and
`f''(2)` is
`+4`, therefore we deduce that the relative maxium is located at
`x=4`, because there the second derivative
`f''(x)` is negative.