Answer
(a) The critical numbers are:
$x = 0$
$x = \frac{4}{7}$
$x = 1$
(b) It is undetermined if $f$ has a local maximum or a local minimum at $x=0$
$f$ has a local minimum at $x=\frac{4}{7}$
It is undetermined if $f$ has a local maximum or a local minimum at $x=1$
(c) $f$ has a local maximum at $x = 0$
$f$ has a local minimum at $x = \frac{4}{7}$
$f$ does not have a local maximum or a local minimum at $x = 1$
Work Step by Step
(a) $f(x) = x^4~(x-1)^3$
We can find the critical numbers:
$f'(x) = 4x^3~(x-1)^3+3~x^4~(x-1)^2 = 0$
$x^3~(x-1)^2~[4~(x-1)+3~x] = 0$
$x^3~(x-1)^2~(7x-4) = 0$
$x = 0, \frac{4}{7}, 1$
The critical numbers are:
$x = 0$
$x = \frac{4}{7}$
$x = 1$
(b) We can use the Second Derivative Test:
$f''(x) = 12x^2~(x-1)^3+12x^3~(x-1)^2+12x^3~(x-1)^2+6~x^4~(x-1)$
$f''(x) = 12x^2~(x-1)^3+24x^3~(x-1)^2+6~x^4~(x-1)$
$f''(x) = 6x^2~(x-1)[2~(x-1)^2+4x~(x-1)+x^2]$
$f''(x) = 6x^2~(x-1)(2x^2-4x+2+4x^2-4x+x^2)$
$f''(x) = 6x^2~(x-1)(7x^2-8x+2)$
$f'(0) = 0$
$f''(0) = 0$
It is undetermined if $f$ has a local maximum or a local minimum at $x=0$
$f'(\frac{4}{7}) = 0$
$f''(\frac{4}{7}) = 1.68 \gt 0$
$f$ has a local minimum at $x=\frac{4}{7}$
$f'(1) = 0$
$f''(1) = 0$
It is undetermined if $f$ has a local maximum or a local minimum at $x=1$
(c) As $x \to 0^-,$ then $f'(x) \gt 0$
As $x \to 0^+,$ then $f'(x) \lt 0$
$f$ has a local maximum at $x = 0$
As $x \to \frac{4}{7}^-,$ then $f'(x) \lt 0$
As $x \to \frac{4}{7}^+,$ then $f'(x) \gt 0$
$f$ has a local minimum at $x = \frac{4}{7}$
As $x \to 1^-,$ then $f'(x) \gt 0$
As $x \to 1^+,$ then $f'(x) \gt 0$
$f$ does not have a local maximum or a local minimum at $x = 1$
