For the function f(x) = 5x^6 + 6x^5 - 15x^4
a.) find where the function is increasing & decreasing
b.) find where the graph if concave up & concave down
c.) name all inflection points
d.) find all relative extrema
To find where the function is increasing and decreasing, we need to find the critical points of the function. The critical points occur where the derivative of the function is equal to zero or is undefined.
a.) Find where the function is increasing and decreasing:
1. Take the derivative of the function f(x) with respect to x:
f'(x) = 30x^5 + 30x^4 - 60x^3
2. Set the derivative equal to zero and solve for x to find the critical points:
30x^5 + 30x^4 - 60x^3 = 0
Factor out 30x^3:
30x^3 (x^2 + x - 2) = 0
From this equation, we have two possibilities:
a) 30x^3 = 0, which gives x = 0 as a critical point.
b) (x^2 + x - 2) = 0, which gives x = -2 and x = 1 as critical points.
Now, we need to determine the intervals in which the function is increasing or decreasing. We do this by looking at the intervals between the critical points and at the values of the derivative in these intervals.
To determine if the function is increasing or decreasing on each interval, plug in a test value from each interval into the derivative f'(x).
Let's pick a value from each interval:
For x < -2, let's take x = -3
For -2 < x < 0, let's take x = -1
For 0 < x < 1, let's take x = 0.5
For x > 1, let's take x = 2
Evaluate f'(x) at each of these test values:
For x = -3:
f'(-3) = 30(-3)^5 + 30(-3)^4 - 60(-3)^3 = 540
For x = -1:
f'(-1) = 30(-1)^5 + 30(-1)^4 - 60(-1)^3 = -60
For x = 0.5:
f'(0.5) = 30(0.5)^5 + 30(0.5)^4 - 60(0.5)^3 = 4.125
For x = 2:
f'(2) = 30(2)^5 + 30(2)^4 - 60(2)^3 = 960
Now we can determine the intervals where the function is increasing or decreasing:
On the interval x < -2, f'(x) = 540 > 0, so f(x) is increasing.
On the interval -2 < x < 0, f'(x) = -60 < 0, so f(x) is decreasing.
On the interval 0 < x < 1, f'(x) = 4.125 > 0, so f(x) is increasing.
On the interval x > 1, f'(x) = 960 > 0, so f(x) is increasing.
b.) To find where the graph of the function is concave up and concave down, we need to find the points of inflection. Points of inflection occur where the second derivative of the function changes sign.
1. Take the second derivative of f(x):
f''(x) = 150x^4 + 120x^3 - 180x^2
2. Set the second derivative equal to zero and solve for x to find the possible points of inflection:
150x^4 + 120x^3 - 180x^2 = 0
Factor out 30x^2:
30x^2 (5x^2 + 4x - 6) = 0
From this equation, we have two possibilities:
a) 30x^2 = 0, which gives x = 0 as a possible point of inflection.
b) (5x^2 + 4x - 6) = 0, which gives x ≈ 0.697 and x ≈ -1.097 as possible points of inflection.
Now, we need to determine if these points are indeed points of inflection. We do this by looking at the sign of the second derivative on each side of these points.
Let's check the sign of f''(x) on both sides of x = 0:
For x < 0, let's take x = -1
For x > 0, let's take x = 1
Evaluate f''(x) at each of these test values:
For x = -1:
f''(-1) = 150(-1)^4 + 120(-1)^3 - 180(-1)^2 = -210
For x = 1:
f''(1) = 150(1)^4 + 120(1)^3 - 180(1)^2 = 90
Since the sign of f''(x) changes from negative to positive at x = 0, x = 0 is a point of inflection.
Similarly, we can check the sign of f''(x) on both sides of x ≈ 0.697 and x ≈ -1.097 to confirm if they are points of inflection.
c.) The inflection points are x = 0, x ≈ 0.697, and x ≈ -1.097.
d.) To find the relative extrema, we look at the critical points obtained in part a.
The critical points are x = 0, x = -2, and x = 1. To determine if these points are relative extrema, we can use the second derivative test.
Evaluate f''(x) at each of the critical points:
For x = 0:
f''(0) = 150(0)^4 + 120(0)^3 - 180(0)^2 = 0
For x = -2:
f''(-2) = 150(-2)^4 + 120(-2)^3 - 180(-2)^2 = 1200
For x = 1:
f''(1) = 150(1)^4 + 120(1)^3 - 180(1)^2 = 90
Using the second derivative test:
- If f''(x) > 0, then f(x) has a local minimum at x.
- If f''(x) < 0, then f(x) has a local maximum at x.
- If f''(x) = 0, the test is inconclusive and we cannot determine if there is a local extremum at x.
Based on the second derivative test:
- At x = 0, the second derivative is 0, so the test is inconclusive.
- At x = -2, the second derivative is positive (f''(-2) = 1200), so there is a local minimum.
- At x = 1, the second derivative is positive (f''(1) = 90), so there is a local minimum.
Therefore, the relative extrema are as follows:
- Local minimum at x = -2
- Local minimum at x = 1