Answer :

We are given the function

$$
y = 12x^5 - 45x^4 + 40x^3.
$$

Our goal is to find its critical points and determine the nature of these points (local maximum, local minimum, or inconclusive) using the first and second derivative tests.

──────────────────────────────
Step 1. Find the First Derivative

Differentiate the function with respect to $x$:

$$
y' = \frac{d}{dx}\left(12x^5 - 45x^4 + 40x^3\right).
$$

Applying the power rule term by term:

\[
\begin{aligned}
y' &= 12 \cdot 5x^{5-1} - 45 \cdot 4x^{4-1} + 40 \cdot 3x^{3-1} \\
&= 60x^4 - 180x^3 + 120x^2.
\end{aligned}
\]

We can factor $60x^2$ from the expression:

$$
y' = 60x^2 \left(x^2 - 3x + 2\right).
$$

Notice that the quadratic factor can be factored further:

$$
x^2 - 3x + 2 = (x-1)(x-2).
$$

Thus, the first derivative becomes:

$$
y' = 60x^2 (x-1)(x-2).
$$

──────────────────────────────
Step 2. Find the Critical Points

Critical points occur when $y' = 0$ or when $y'$ is undefined. Here, $y'$ is defined for all $x$, so we set

$$
60x^2 (x-1)(x-2) = 0.
$$

Solving for $x$, we have:

- From $60x^2 = 0$, we get $x = 0$.
- From $x-1 = 0$, we get $x = 1$.
- From $x-2 = 0$, we get $x = 2$.

Thus, the critical points are $x = 0$, $x = 1$, and $x = 2$.

──────────────────────────────
Step 3. Evaluate the Function $y$ at Critical Points

We now calculate the function values at these critical points.

1. For $x=0$:

$$
y(0) = 12(0)^5 - 45(0)^4 + 40(0)^3 = 0.
$$

2. For $x=1$:

$$
y(1) = 12(1)^5 - 45(1)^4 + 40(1)^3 = 12 - 45 + 40 = 7.
$$

3. For $x=2$:

\[
\begin{aligned}
y(2) &= 12(2)^5 - 45(2)^4 + 40(2)^3 \\
&= 12 \cdot 32 - 45 \cdot 16 + 40 \cdot 8 \\
&= 384 - 720 + 320 \\
&= -16.
\end{aligned}
$$

──────────────────────────────
Step 4. Find the Second Derivative

Differentiate $y'$ to obtain the second derivative $y''$:

$$
y' = 60x^4 - 180x^3 + 120x^2.
$$

Differentiate term by term:

\[
\begin{aligned}
y'' &= \frac{d}{dx}\left(60x^4\right) - \frac{d}{dx}\left(180x^3\right) + \frac{d}{dx}\left(120x^2\right) \\
&= 60 \cdot 4x^{3} - 180 \cdot 3x^{2} + 120 \cdot 2x \\
&= 240x^3 - 540x^2 + 240x.
\end{aligned}
\]

──────────────────────────────
Step 5. Classify Each Critical Point Using the Second Derivative Test

Evaluate $y''$ at each critical point.

1. At $x = 0$:

$$
y''(0) = 240(0)^3 - 540(0)^2 + 240(0) = 0.
$$

Since $y''(0) = 0$, the second derivative test is inconclusive at $x = 0$. This point might be a saddle point or a point of inflection.

2. At $x = 1$:

$$
y''(1) = 240(1)^3 - 540(1)^2 + 240(1) = 240 - 540 + 240 = -60.
$$

Because $y''(1) < 0$, the function is concave down at $x = 1$, indicating a local maximum. Thus, there is a local maximum at $(1, 7)$.

3. At $x = 2$:

$$
y''(2) = 240(2)^3 - 540(2)^2 + 240(2) = 240 \cdot 8 - 540 \cdot 4 + 480 = 1920 - 2160 + 480 = 240.
$$

Because $y''(2) > 0$, the function is concave up at $x = 2$, indicating a local minimum. Thus, there is a local minimum at $(2, -16)$.

──────────────────────────────
Step 6. Summary of Extreme Values

- The critical points are $x = 0$, $x = 1$, and $x = 2$.
- At $x = 0$, $y(0)=0$, and the second derivative test is inconclusive.
- At $x = 1$, $y(1)=7$, and the second derivative is negative ($y''(1)=-60$), so there is a local maximum at $\boxed{(1, 7)}$.
- At $x = 2$, $y(2)=-16$, and the second derivative is positive ($y''(2)=240$), so there is a local minimum at $\boxed{(2, -16)}$.

──────────────────────────────
Final Answer

The function has the following extreme values:

- A local maximum at $(1, 7)$.
- A local minimum at $(2, -16)$.
- The point $(0, 0)$ is a critical point with an inconclusive result from the second derivative test.

This completes the step-by-step solution.

Other Questions