We are given the function:

$f(x) = x^5 - 4x^2 + 1$

We need to evaluate $f(-1)$ and $f(2)$.

1. Finding $f(-1)$:

Substitute $x = -1$ into the function:

$f(-1) = (-1)^5 - 4(-1)^2 + 1$ $f(-1) = -1 - 4(1) + 1$ $f(-1) = -1 - 4 + 1 = -4$

2. Finding $f(2)$:

Substitute $x = 2$ into the function:

$f(2) = 2^5 - 4(2^2) + 1$ $f(2) = 32 - 4(4) + 1$ $f(2) = 32 - 16 + 1 = 17$

Final Results:

$f(-1) = -4, \quad f(2) = 17$

Would you like more details on how these calculations were done, or do you have any questions?

Related Questions:

1. What is the derivative of $f(x) = x^5 - 4x^2 + 1$?
2. Can we find the roots of the equation $f(x) = 0$?
3. How would you find the local extrema of $f(x)$?
4. What is the behavior of $f(x)$ as $x \to \infty$ and $x \to -\infty$?
5. How does the function change if we add a constant to it?

Tip:

When substituting negative numbers into exponents, remember that even exponents will make the result positive, while odd exponents will preserve the negative sign.