Question

given the function, $f(x)=x^{3}-4x^{2}+x - 4$, what are the coordinates of the relative minimum? (round your answer to the nearest 10 place)

Explanation:

Step 1: Find the first derivative

To find the critical points, we first find the derivative of \( f(x) = x^3 - 4x^2 + x - 4 \). Using the power rule, the derivative \( f'(x) \) is:
\( f'(x) = 3x^2 - 8x + 1 \)

Step 2: Solve \( f'(x) = 0 \)

We solve the quadratic equation \( 3x^2 - 8x + 1 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -8 \), and \( c = 1 \).
\[

$$\begin{align*} x &= \frac{8 \pm \sqrt{(-8)^2 - 4(3)(1)}}{2(3)} \\ &= \frac{8 \pm \sqrt{64 - 12}}{6} \\ &= \frac{8 \pm \sqrt{52}}{6} \\ &= \frac{8 \pm 2\sqrt{13}}{6} \\ &= \frac{4 \pm \sqrt{13}}{3} \end{align*}$$

\]
Calculating the numerical values:
\( \sqrt{13} \approx 3.6055 \)
\( x_1 = \frac{4 + 3.6055}{3} \approx \frac{7.6055}{3} \approx 2.535 \)
\( x_2 = \frac{4 - 3.6055}{3} \approx \frac{0.3945}{3} \approx 0.1315 \)

Step 3: Find the second derivative

To determine if these critical points are minima or maxima, we find the second derivative \( f''(x) \):
\( f''(x) = 6x - 8 \)

Step 4: Test the critical points

• For \( x \approx 0.1315 \):

\( f''(0.1315) = 6(0.1315) - 8 \approx 0.789 - 8 = -7.211 < 0 \), so this is a relative maximum.

• For \( x \approx 2.535 \):

\( f''(2.535) = 6(2.535) - 8 \approx 15.21 - 8 = 7.21 > 0 \), so this is a relative minimum.

Step 5: Find the corresponding \( y \)-coordinate

Now we find \( f(2.535) \):
\[

$$\begin{align*} f(2.535) &= (2.535)^3 - 4(2.535)^2 + 2.535 - 4 \\ &\approx 16.29 - 4(6.426) + 2.535 - 4 \\ &\approx 16.29 - 25.704 + 2.535 - 4 \\ &\approx -10.879 \end{align*}$$

\]

Answer:

\((2.54, -10.88)\) (rounded to the nearest hundredth, assuming the problem meant nearest hundredth as "nearest 10 place" might be a typo; if nearest tenth, it would be \((2.5, -10.9)\))