The function f(x) = x9 + 3x7 + 64 is increasing on

To determine where the function is increasing, we need to find its derivative and set it equal to zero to find the critical points.

The derivative of f(x) = x^9 + 3x^7 + 64 is f'(x) = 9x^8 + 21x^6.

Setting f'(x) equal to zero gives us:

9x^8 + 21x^6 = 0

We can factor out 3x^6 from the equation:

3x^6(3x^2 + 7) = 0

Setting each factor equal to zero gives us the critical points x = 0 and x = -√(7/3), x = √(7/3).

To determine where the function is increasing, we test the intervals between the critical points in the derivative.

For x < -√(7/3), f'(x) > 0, so the function is increasing.

For -√(7/3) < x < 0, f'(x) < 0, so the function is decreasing.

For 0 < x < √(7/3), f'(x) > 0, so the function is increasing.

For x > √(7/3), f'(x) > 0, so the function is increasing.

Therefore, the function f(x) = x^9 + 3x^7 + 64 is increasing on the intervals (-∞, -√(7/3)) U (0, √(7/3)) U (√(7/3), ∞).