Find the critical numbers, increasing intervals, x-values where f has an inflection point, and where the graph is concave down from the equation f(x)=2x^4-3x^3-10x^2+20
To find the critical numbers, we need to find the values of x where the derivative of the function f(x) equals zero or is undefined.
Step 1: Calculate the derivative of f(x):
f'(x) = 8x^3 - 9x^2 - 20x
Step 2: Set f'(x) equal to zero and solve for x:
8x^3 - 9x^2 - 20x = 0
There are multiple ways to solve this equation. One way is to factor out x:
x(8x^2 - 9x - 20) = 0
Applying the zero product property, we can set each factor equal to zero:
x = 0 (one critical number)
Solving the quadratic equation, we get two more critical numbers:
8x^2 - 9x - 20 = 0
(x - 2)(8x + 10) = 0
x - 2 = 0 or 8x + 10 = 0
x = 2 or x = -10/8 = -5/4 (two more critical numbers)
Therefore, the critical numbers are x = 0, x = 2, and x = -5/4.
To determine the increasing intervals, we can use the first derivative test. We need to evaluate the sign of f'(x) on different intervals.
Step 3: Choose test points in each interval and evaluate f'(x) at these points.
Let's test f'(x) at x = -1, x = 1, and x = 3:
f'(-1) = 8(-1)^3 - 9(-1)^2 - 20(-1) = 8 + 9 - 20 = -3
f'(1) = 8(1)^3 - 9(1)^2 - 20(1) = 8 - 9 - 20 = -21
f'(3) = 8(3)^3 - 9(3)^2 - 20(3) = 216 - 81 - 60 = 75
Step 4: Analyze the sign of f'(x) in each interval:
For x < -5/4, f'(x) is positive (+) since f'(-1) = -3 < 0
For -5/4 < x < 0, f'(x) is negative (-) since f'(-1) = -3 < 0 and f'(1) = -21 < 0
For 0 < x < 2, f'(x) is positive (+) since f'(1) = -21 < 0 and f'(3) = 75 > 0
For x > 2, f'(x) is negative (-) since f'(3) = 75 > 0
Therefore, the increasing intervals are (-∞, -5/4) and (0, 2).
To find the x-values where f has an inflection point and where the graph is concave down, we need to find the values of x where the second derivative of f(x) equals zero or is undefined.
Step 5: Calculate the second derivative of f(x):
f''(x) = 24x^2 - 18x - 20
Step 6: Set f''(x) equal to zero and solve for x:
24x^2 - 18x - 20 = 0
There are multiple ways to solve this equation. One way is to use the quadratic formula:
x = (-b ± √(b^2 - 4ac))/(2a)
Using this formula, we get:
x = (18 ± √(18^2 - 4(24)(-20)))/(2(24))
x = (18 ± √(324 + 1920))/48
x = (18 ± √2244)/48
Simplifying further, we have:
x = (18 ± 47.38)/48
x = (18 + 47.38)/48 or x = (18 - 47.38)/48
x = 65.38/48 or x = -29.38/48
x ≈ 1.36 or x ≈ -0.61
Therefore, the x-values where f has an inflection point are x ≈ 1.36 and x ≈ -0.61.
To determine where the graph is concave down, we need to evaluate the sign of f''(x) on each side of the inflection points.
Step 7: Choose test points in each interval and evaluate f''(x) at these points.
Let's test f''(x) at x = 0, x = 1, and x = 2:
f''(0) = 24(0)^2 - 18(0) - 20 = -20
f''(1) = 24(1)^2 - 18(1) - 20 = 24 - 18 - 20 = -14
f''(2) = 24(2)^2 - 18(2) - 20 = 96 - 36 - 20 = 40
Step 8: Analyze the sign of f''(x) in each interval:
For x < -0.61, f''(x) is negative (-) since f''(0) = -20 < 0
For -0.61 < x < 1.36, f''(x) is positive (+) since f''(0) = -20 < 0 and f''(1) = -14 < 0
For x > 1.36, f''(x) is positive (+) since f''(1) = -14 < 0 and f''(2) = 40 > 0
Therefore, the graph is concave down on (-∞, -0.61) and (1.36, ∞).
To summarize:
- Critical numbers: x = 0, x = 2, and x = -5/4
- Increasing intervals: (-∞, -5/4) and (0, 2)
- x-values with inflection points: x ≈ 1.36 and x ≈ -0.61
- Graph concave down: (-∞, -0.61) and (1.36, ∞)