Answer:
f is increasing on (3, infinity)
Explanation:
On which interval is the function f(x) = x^2 - 6x + 4 increasing?
1) Here's an approach that doesn't involve calculus:
Determine the vertex of this parabolic graph. The x-coordinate of the vertex is x = -b / (2a), which here is x = -(-6) / (2*1), or x = 6/2, or x = 3.
The y-coord. of the vertex is f(3), which comes out to 3^2 - 6(3) + 4 = -5. The vertex is then (3, -5). Because the coeff. of the x^2 term is +, we know that the parabola opens up. From x = -infinity to x = 3, the function is decreasing; from x = 3 to inf, the function is increasing.
2) Here's the calculus approach: The derivative of f(x) = x^2 - 6x + 4 is
f '(x) = 2x - 6. We set this = to 0 and solve for x: x = 3.
This tells us that the x-coordinate of the vertex is 3 (which we already knew).
Choose two test points: x = 2 (on the left of x = 3) and x = 4 (on the right of x = 3). Substitute 2 for x in f '(x) = 2x - 6; we get a negative result, which tells us that the function is decreasing on (-inf., 3). Subst. 4 for x in f '(x) = 2x - 6; we get a positive result, which tells us that f is increasing on (3, infinity).