for f(x= -1/4x4-1/3x3+2x, use analytic method to find the exact intervals on which the function is a)increasing, b)decreasing, c)concave up, d)concave down, then find any e)local extreme values, f)inflection points
y = -1/4 x^4 - 1/3 x^3 + 2x
y' = -x^3 - x^2 + 2
y'' = -3x^2 - 2x
inflection where y''=0: x=0, -2/3
local extrema where y'=0 and y''≠0: x=1
increasing where y'>0: x<1
decreasing where y'<0: x>1
concave up where y''>0: -2/3 < x < 0
concave down elsewhere
how did you factor y' for this question?
To find the intervals on which a function is increasing, decreasing, concave up, concave down, and the local extreme values and inflection points, we need to follow a systematic approach:
a) Increasing intervals:
To find where the function is increasing, we need to determine where the derivative is positive.
1. Take the derivative of the function:
f'(x) = -x^3 + x^2 + 2
2. Solve for f'(x) > 0:
-x^3 + x^2 + 2 > 0
To solve this inequality, we can use different methods such as graphing, factoring, or the quadratic formula. In this case, we can try factoring and analyzing the sign of each factor:
-x^3 + x^2 + 2 = (x + 1)(-x - 1)(x - 2) > 0
Now, analyze the sign of each factor on the number line:
- - + - + +
-1 0 1 2 3 4
Thus, f'(x) > 0 for x ∈ (-∞, -1) ∪ (2, ∞). These are the intervals where the function is increasing.
b) Decreasing intervals:
To find where the function is decreasing, we need to determine where the derivative is negative.
From earlier, we know that f'(x) = -x^3 + x^2 + 2.
To solve f'(x) < 0, we can analyze the sign of each factor again:
+ - + - + -
-1 0 1 2 3 4
Thus, f'(x) < 0 for x ∈ (-1, 0) U (1, 2). These are the intervals where the function is decreasing.
c) Concave up intervals:
To find where the function is concave up, we need to determine where the second derivative is positive.
1. Take the second derivative of the function:
f''(x) = -3x^2 + 2x
2. Solve for f''(x) > 0:
-3x^2 + 2x > 0
To solve this inequality, we can factor:
-x(-3x + 2) > 0
Now, analyze the sign of each factor on the number line:
+ - +
0 1/3 2
Thus, f''(x) > 0 for x ∈ (0, 2). These are the intervals where the function is concave up.
d) Concave down intervals:
To find where the function is concave down, we need to determine where the second derivative is negative.
Using f''(x) = -3x^2 + 2x, we can analyze the sign of each factor:
- + -
0 1/3 2
Thus, f''(x) < 0 for x ∈ (-∞, 0) ∪ (1/3, 2). These are the intervals where the function is concave down.
e) Local extreme values:
To find local extreme values, we need to locate the critical points where the derivative is either zero or undefined.
For this function, the derivative f'(x) = -x^3 + x^2 + 2.
To find critical points, we set f'(x) = 0, and solve for x:
-x^3 + x^2 + 2 = 0
To find the exact values of x, you might need to use numerical methods or a graphing calculator.
f) Inflection points:
To find the inflection points, we need to locate the points where the concavity changes.
For this function, the second derivative f''(x) = -3x^2 + 2x.
To find the inflection points, we set f''(x) = 0, and solve for x:
-3x^2 + 2x = 0
Use factoring or the quadratic formula to solve for the exact values of x.
Note: Due to the complexity of the equation, the exact solutions may require numerical methods or a graphing calculator.