Determine the critical values for the function y = (1/3)x^3 + x^2 - 8x + 1, and give the smaller value first.

Step-by-step solution

1. 1

   Know what a critical value is before differentiating. Critical values are the $x$ where $y'=0$ or where $y'$ fails to exist. This $y$ is a polynomial, so its derivative exists everywhere and the second possibility drops out — the whole problem is solving $y'=0$. The $\tfrac13$ in front of $x^{3}$ is there precisely so the derivative comes out with a leading coefficient of $1$.

2. 2

   Differentiate term by term with the power rule. Each term is a power of $x$ times a constant:

   $y'=\frac{1}{3}\cdot 3x^{2}+2x-8+0=x^{2}+2x-8$

   The constant $+1$ contributes nothing, as every constant does.

3. 3

   Set the derivative to zero. Critical values are the solutions of

   $x^{2}+2x-8=0$

4. 4

   Factor rather than reaching for the quadratic formula. Look for two numbers whose product is $-8$ and whose sum is $+2$; those are $+4$ and $-2$:

   $x^{2}+2x-8=(x+4)(x-2)=0$

5. 5

   Solve each factor and order the answers. Setting each factor to zero gives

   $x+4=0\ \Rightarrow\ x=-4,\qquad x-2=0\ \Rightarrow\ x=2$

   So the critical values are $x=-4$ and $x=2$, and the smaller is $x=-4$.

6. 6

   Confirm both the derivative and the roots. Substituting back, $(-4)^{2}+2(-4)-8=16-8-8=0$ and $2^{2}+2(2)-8=4+4-8=0$, so both are genuine roots. A central-difference derivative of the original $y$ gives $-6.75$ at $x=0.5$ and $-8.96$ at $x=-1.2$, matching $x^{2}+2x-8$ at those points, so the derivative itself is right.