Give the interval(s) on which the function f is positive if f is defined as:


f(x)=x(5−x)(2x−3)

Try writing it as

f(x) = -x(x-5)(2x-3)

Now you can see that the coefficient of x^3 is negative.

So, for x very large negative, say, -100, f(x) is positive. So, the graph comes down on the left, crosses the x axis at 0,3/2,5, and then continues on down to the right.

So, f(x) > 0 on (-∞,0)U(3/2,5)

To determine the intervals on which the function f(x) = x(5 - x)(2x - 3) is positive, we need to find the values of x that make the function greater than zero.

First, let's identify the critical points of the function. These are the values of x where the function could potentially change sign. The critical points occur when the function is equal to zero or undefined.

Setting f(x) to zero, we have:

x(5 - x)(2x - 3) = 0

Since a product is zero when at least one of its factors is zero, we can set each factor equal to zero individually:

x = 0, 5 - x = 0 (which gives x = 5),
and 2x - 3 = 0 (which gives x = 3/2).

So, the critical points of f(x) are x = 0, x = 5, and x = 3/2.

Next, we create a sign chart by dividing the number line into intervals separated by the critical points. We evaluate the function in each interval to determine its sign.

Interval 1: (-∞, 0)
Pick any value in this interval, e.g., x = -1.
f(-1) = (-1)(5 - (-1))(2(-1) - 3) = (-1)(6)(-5) = 30 > 0

Interval 2: (0, 3/2)
Pick any value in this interval, e.g., x = 1.
f(1) = (1)(5 - 1)(2(1) - 3) = (1)(4)(-1) = -4 < 0

Interval 3: (3/2, 5)
Pick any value in this interval, e.g., x = 4.
f(4) = (4)(5 - 4)(2(4) - 3) = (4)(1)(5) = 20 > 0

Interval 4: (5, ∞)
Pick any value in this interval, e.g., x = 6.
f(6) = (6)(5 - 6)(2(6) - 3) = (6)(-1)(9) = -54 < 0

Based on the sign chart, the function f(x) is positive on the intervals (-∞, 0) and (3/2, 5).

Therefore, the function f(x) = x(5 - x)(2x - 3) is positive for x in the intervals (-∞, 0) and (3/2, 5).