a.) Determine the critical numbers of $f(x) = x^4 ( x- 1 )^3.$

b.) What does the Second Derivative Test tell you about the behaviour of $f$ at these critical numbers?

c.) What does First Derivative Test tell you?

a.) If $f(x) = x^4(x-1)^3$, then by using Product Rule and Chain Rule.

$ \begin{equation} \begin{aligned} f'(x) &= x^4 \left(3(x-1)^2 \right) + 4x^3 (x-1)^3 = 3x^4(x-1)^2 + 4x^3 (x-1)^3\\ \\ f''(x) &= 12x^3 (x-1)^2 + 3x^4(2(x-1)) + 12x^2 ( x- 1 )^3 + 4x^3 (3(x-1)^2)\\ \\ f''(x) &= 24x^3(x-1)^2 + 6x^4 (x-1) + 12x^2 (x-1)^3 \end{aligned} \end{equation} $

Which can be simplified as,

$f''(x) = 6x^2 \left[ 7x^3 - 15x^2 + 10 x - 2 \right]$

To determine the critical numbers, we set $f'(x) = 0$, so...

$ \begin{equation} \begin{aligned} f'(x) = 0 &= 3x^4 (x -1)^2 + 4x^3(x-1)^3\\ \\ 0 &= x^3(x-1)^2 \left[ 3x + 4 (x-1) \right] \end{aligned} \end{equation} $

We have,

$ x^3 = 0, \quad (x-1)^2 = 0, \quad $ and $\quad 3x + 4 (x-1) = 0$

Therefore, the critical numbers are, $x =0$, $x= 1$ and $\displaystyle x = \frac{4}{7}$

If we evaluate $f''(x)$ at the critical numbers, we get...

$ \begin{equation} \begin{aligned} \text{when } x &= 0,\\ \\ f''(0) &= 6(0)^2 \left[ 7(0)^3 - 15 (0)^2 + 10(0) - 2\right]\\ \\ f''(0) &= 0\\ \\ \text{when } x &= 1,\\ \\ f''(1) &= 6(1)^2 \left[ 7(1)^3 - 15 (1)^2 + 10(1) - 2\right]\\ \\ f''(1) &= 0\\ \\ \text{when } x &= \frac{4}{7},\\ \\ f''\left(\frac{4}{7}\right) &= 6\left(\frac{4}{7}\right)^2 \left[ 7\left(\frac{4}{7}\right)^3 - 15 \left(\frac{4}{7}\right)^2 + 10\left(\frac{4}{7}\right) - 2\right]\\ \\ f''\left(\frac{4}{7}\right) &= 0 \end{aligned} \end{equation} $

Since $f''(1) = f''(0) = 0$. It means that $x = 0$ and $x = 1 $ are inflection points and since $\displaystyle f''\left(\frac{4}{7}\right) > 0$, it means that $\displaystyle x = \frac{4}{7}$ is a local minimum.

c.) By using First Derivative Test, if we divide $f$ on the interval:

$ \begin{array}{|c|c|c|} \hline\\ \text{Interval} & f' & f\\ \hline\\ x < 0 & + & \text{increasing on } (-\infty, 0)\\ \hline\\ 0 < x < \frac{4}{7} & - & \text{decreasing on } \left(0,\frac{4}{7}\right)\\ \hline\\ \frac{4}{7} < x < 1 & + & \text{increasing on } \left(\frac{4}{7},1\right)\\ \hline\\ x > 1 & + & \text{increasing on } (1, \infty)\\ \hline \end{array} $

Since the function changes from positive to negative at $x = 0$, we can say that 0 is a local maximum. Also, we have $\displaystyle \frac{4}{7}$ as a local minimum since the function changes from negative to positive at the point. We can see in the table that 1 is an inflection point since the sign of $f'$ did not vary.

Notice that the answer in part(b) differs from part(a) because of the repeating roots of $f'(x)$.

Posted on

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now