f(x) = x^3 – 25x (b) Find the domain and range of f. (Enter your answers using interval notation.) Domain: (-∞, ∞) Range: (-∞, ∞) (c) State approximately the intervals on which f is increasing and on which f is decreasing. (Enter your answers using interval notation. Round your answers to two decimal places.) Increasing: (-∞, -2.92) ∪ (2.92, ∞) Decreasing: (-2.92, 2.92)
The Correct Answer and Explanation is:
Let us analyze the given function:f(x)=x3−25xf(x) = x^3 – 25xf(x)=x3−25x
Part (b): Domain and Range
Domain:
The function is a polynomial. Polynomials are defined for all real numbers because there are no restrictions such as division by zero or taking the square root of a negative number. Therefore, the domain is all real numbers.Domain: (−∞,∞)\text{Domain: } (-\infty, \infty)Domain: (−∞,∞)
Range:
Similarly, the range of a cubic function that includes both odd and linear powers will also cover all real numbers. As xxx approaches positive or negative infinity, f(x)f(x)f(x) also approaches positive or negative infinity respectively.Range: (−∞,∞)\text{Range: } (-\infty, \infty)Range: (−∞,∞)
Part (c): Intervals of Increase and Decrease
To find where the function is increasing or decreasing, the first derivative is used. The first derivative represents the slope of the tangent line to the function.f′(x)=3×2−25f'(x) = 3x^2 – 25f′(x)=3×2−25
Set the derivative equal to zero to find the critical points:3×2−25=03x^2 – 25 = 03×2−25=0
Solving for xxx:x2=253x^2 = \frac{25}{3}x2=325x=±253≈±2.89x = \pm \sqrt{\frac{25}{3}} \approx \pm 2.89x=±325≈±2.89
Now, test the intervals determined by these critical points:
- For x<−2.89x < -2.89x<−2.89, choose x=−3x = -3x=−3:
f′(−3)=3(−3)2−25=27−25=2 (positive)f'(-3) = 3(-3)^2 – 25 = 27 – 25 = 2 \, (positive)f′(−3)=3(−3)2−25=27−25=2(positive)
The function is increasing on (−∞,−2.89)(-\infty, -2.89)(−∞,−2.89).
- For −2.89<x<2.89-2.89 < x < 2.89−2.89<x<2.89, choose x=0x = 0x=0:
f′(0)=−25 (negative)f'(0) = -25 \, (negative)f′(0)=−25(negative)
The function is decreasing on (−2.89,2.89)(-2.89, 2.89)(−2.89,2.89).
- For x>2.89x > 2.89x>2.89, choose x=3x = 3x=3:
f′(3)=3(3)2−25=27−25=2 (positive)f'(3) = 3(3)^2 – 25 = 27 – 25 = 2 \, (positive)f′(3)=3(3)2−25=27−25=2(positive)
The function is increasing on (2.89,∞)(2.89, \infty)(2.89,∞).
Correct Answer (rounded to two decimal places):
Increasing:(−∞,−2.89)∪(2.89,∞)(-\infty, -2.89) \cup (2.89, \infty)(−∞,−2.89)∪(2.89,∞)
Decreasing:(−2.89,2.89)(-2.89, 2.89)(−2.89,2.89)
This analysis shows that the function has one local maximum and one local minimum. The function increases before and after the critical points and decreases between them.
