0

Here's my equation:

$y=x(x-4)^{3}$

I'm supposed to find the increasing and decreasing intervals, which I know how to do for other problems but this one is giving me issues.

My first question is: should I expand the equation before taking the derivative? Because so far I've tried taking the derivative as-is and I get stuck.

${y}'= {x}'[(x-4)^{3}]+x{[(x-4)^{3}]}'$
${y}'= (x-4)^{3}+3x[(x-4)^{2}]$

... and that's as far as I understand. However, a study website I'm using suggests this as the next step:

${y}'= (x-4)^{2}(4x-4)$
$0= (x-4)^{2}(4x-4)$
$x=4, x=1$

None of that makes sense to me. I don't see how they simplified it down, or how they pulled out $x=4, x=1$.

So I'm wondering if there's a better way, perhaps expanding the original equation before taking the derivative? I don't really know though and I'm mostly looking for some direction. I understand how to use the critical points to find the intervals, I just don't get how to find the critical points here.

Grav
  • 133

1 Answers1

1

You're so close! The derivative is best written like so $$ \begin{split} f'(x) =& (x-4)^3 + x\cdot 3(x-4)^2 \\ =& (x-4)^2 \left[ (x-4) + 3x \right] \\ =& (x-4)^2 \left[ 4x-4 \right] \end{split} $$ So when is this zero? The expression is zero when $$ (x-4)^2 = 0 \qquad \text{or} \qquad [4x-4]= 0 $$ The solutions are easy to see, $x=4$ or $x=1$, respectively. Looking at the values of $f'(x)$ around and between these two values, we see that

$$ \begin{split} f'(x) < 0 \qquad &\text{when} \qquad x<1 \\ f'(x) > 0 \qquad &\text{when} \qquad 1<x<4 \\ f'(x) > 0 \qquad &\text{when} \qquad 4<x \\ \end{split} $$

Matti P.
  • 6,012