## Calculus: Early Transcendentals 8th Edition

(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 $x = 1$
(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 $x = 1$