The polynomial provided in the image is:

$f(x) = 15x^5 - 5x^4 + 40x^3 + 6$

To find the maximum and minimum values of this function, we analyze its critical points by following these steps:

1. Take the derivative of the function:

The derivative of $f(x)$ will help us identify the critical points where the slope is zero.

$f'(x) = 75x^4 - 20x^3 + 120x^2$

2. Factorize the derivative:

Factorizing $f'(x)$ simplifies finding the critical points:

$f'(x) = 5x^2(15x^2 - 4x + 24)$

3. Solve $f'(x) = 0$:

Set $f'(x) = 0$:

$5x^2(15x^2 - 4x + 24) = 0$

This gives two parts:

• $5x^2 = 0 \implies x = 0$
• $15x^2 - 4x + 24 = 0$

The quadratic equation $15x^2 - 4x + 24 = 0$ has a discriminant:

$\Delta = (-4)^2 - 4(15)(24) = 16 - 1440 = -1424$

Since $\Delta < 0$, the quadratic has no real roots. Therefore, the only critical point is:

$x = 0$

4. Evaluate $f(x)$ at critical points and endpoints (if any):

At $x = 0$:

$f(0) = 15(0)^5 - 5(0)^4 + 40(0)^3 + 6 = 6$

Since the function is a polynomial of degree 5 (odd degree), it tends to $-\infty$ as $x \to -\infty$ and $+\infty$ as $x \to +\infty$. Thus, there are no global maximums or minimums.

The critical point at $x = 0$ gives a local minimum of $f(x) = 6$.

Summary:

• Local minimum: $f(x) = 6$ at $x = 0$.
• No global maximum or minimum as the polynomial grows unbounded.

Would you like a graph or further explanation of this process? 😊

Related Questions:

1. What are the second derivative and its role in identifying maxima or minima?
2. How can we confirm the local behavior using a graph?
3. Why does the polynomial's degree affect its behavior at infinity?
4. What happens if the derivative has multiple real roots?
5. How is the discriminant used to determine real roots of a quadratic?

Tip:

For higher-degree polynomials, plotting the function can provide helpful visual insights into the behavior of critical points.