#### For the function f(x)=45x³−3x^{5}, determine the interval(s) for which f(x) decreases.

**Final Answer:**

The function f(x)=45x^{3}−3x^{5 }decreases on the intervals (−∞,−3) and (3,∞). This was determined by calculating the derivative and testing the sign of the derivative across the critical points. The intervals were confirmed through sign evaluation around the critical points found from the derivative calculation.

**Examples & Evidence:**

To determine the intervals where the function f(x)=45x^{3}−3x^{5 }decreases, we begin by finding its derivative. The derivative f′(x) gives us the rate of change of the function, which helps us identify intervals of increase and decrease.

1. First, we calculate the derivative:

f′(x)=135x^{2}−15x^{4}

This simplifies to:

f′(x)=15x^{2}(9−x^{2})

2. Next, we need to find the critical points where the derivative is equal to zero:

15x^{2}(9−x^{2})=0

Setting each factor to zero gives:

15x^{2}=0⟹x=0

9−x^{2}=0⟹x=±3

So the critical points are x=−3,0,3.

3. Now, we will test intervals defined by these critical points to determine where the derivative is positive (increasing) or negative (decreasing). The intervals to test are:

- (−∞,−3)
- (−3,0)
- (0,3)
- (3,∞)

4. Choose a test point from each interval:

- For x=−4 (in (−∞,−3)):

f′(−4)=15(−4)²(9−(−4)²)=15(16)(9−16)<0(decreasing) - For x=−1 (in (−3,0)):

f′(−1)=15(−1)²(9−1)=15(1)(8)>0

(increasing) - For x=1 (in (0,3)):

f′(1)=15(1)²(9−1)=15(1)(8)>0

(increasing) - For x=4 (in (3,∞)):

f′(4)=15(4)²(9−16)=15(16)(−7)<0

(decreasing)

5. From this testing, we see that:

- The function is decreasing on the intervals (−∞,−3) and (3,∞).
- The function is increasing on the intervals (−3,0) and (0,3).

Thus, the function f(x) decreases in the intervals (−∞,−3) and (3,∞).

**Explanation:**

For example, if we check a point like x=−4, we find that the function decreases because the derivative at that point is negative. Similarly, for x=4, it also indicates the function decreases in that interval, confirming our findings.

The findings are supported by critical point analysis where the derivative changes signs at the points −3,0, and 3. This reveals increasing and decreasing behavior relative to these critical points.

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