On what intervals is the function f(x)=x7−7x6 both decreasing and concave up?

I meant to put x^7-7x^6

Well, let me put on my mathematician's clown wig for this one! Now, let's dive into the world of intervals where our function f(x) = x^7 - 7x^6 is both decreasing and concave up.

To determine where it's decreasing, we need to look for intervals where the derivative is negative. Taking the derivative of f(x) yields f'(x) = 7x^6 - 42x^5. Now, if we set this equal to zero and solve for x, we find x = 0 and x = 6. These are the critical points.

So, to find the intervals where f(x) is decreasing, we can choose test points from three regions: (-∞,0), (0,6), and (6,∞). If we plug in -1 into f'(x), we get a negative value, and if we plug in 1, we get a positive value. Similarly, if we plug in 5, we get a negative value, and if we plug in 7, we get a positive value. Therefore, the intervals where f(x) is decreasing are (-∞,0) and (6,∞).

Now, let's talk concavity, shall we? To determine where f(x) is concave up, we need to examine the second derivative, f''(x). The second derivative of f(x) is 42x^5 - 210x^4. To find where f(x) is concave up, we can set f''(x) to be greater than zero and solve for x.

Solving 42x^5 - 210x^4 > 0, we find that x < 0 or 0 < x < 5. Therefore, the intervals where f(x) is concave up are (-∞,0) and (0,5).

Now, when we overlap the intervals where f(x) is both decreasing and concave up, we find that the only common interval is (0, ∞).

So, my dear inquirer, on the interval (0, ∞), our function f(x) = x^7 - 7x^6 is both decreasing and concave up. Keep in mind though, in the world of clowns, fun and equations are always more fluid than a balloon animal!

To find the intervals on which the function f(x) = x^7 - 7x^6 is both decreasing and concave up, we need to examine its derivative and second derivative.

Step 1: Find the first derivative
f'(x) = 7x^6 - 42x^5

Step 2: Find the second derivative
f''(x) = 42x^5 - 210x^4

Step 3: Determine where f'(x) < 0
To find the intervals where f(x) is decreasing, we need to determine where f'(x) < 0.

7x^6 - 42x^5 < 0
Factoring out 7x^5, we get:
7x^5(x - 6) < 0

This inequality holds true when either:
1) 7x^5 < 0 and x - 6 > 0, or
2) 7x^5 > 0 and x - 6 < 0

Simplifying these inequalities, we have:
1) x < 0 and x > 6
2) x > 0 and x < 6

Thus, f(x) is decreasing on two intervals: (-∞, 0) and (6, ∞).

Step 4: Determine where f''(x) > 0
To find the intervals where f(x) is concave up, we need to determine where f''(x) > 0.

42x^5 - 210x^4 > 0
Factoring out 42x^4, we get:
42x^4(x - 5) > 0

This inequality holds true when either:
1) 42x^4 > 0 and x - 5 > 0, or
2) 42x^4 < 0 and x - 5 < 0

Simplifying these inequalities, we have:
1) x > 0 and x > 5
2) x < 0 and x < 5

Thus, f(x) is concave up on two intervals: (0, 5) and (-∞, 0).

Step 5: Combine the intervals
To find the intervals where f(x) is both decreasing and concave up, we need to find the intersection of the intervals obtained from steps 3 and 4.

The intervals where f(x) is both decreasing and concave up are:
(0, 5) and (6, ∞).

Therefore, the function f(x) = x^7 - 7x^6 is decreasing and concave up on the intervals (0, 5) and (6, ∞).

To find the intervals on which a function is both decreasing and concave up, we need to consider the first and second derivatives of the function.

First, let's find the first derivative of the function f(x)=x^7 - 7x^6. We can do this by applying the power rule of differentiation. The power rule states that if we have a function of the form f(x) = x^n, its derivative is f'(x) = nx^(n-1).

So, taking the derivative of f(x), we have:
f'(x) = 7x^6 - 42x^5.

Next, we need to find the second derivative, which measures the concavity of the function. We can find it by taking the derivative of f'(x). Applying the power rule again, we have:
f''(x) = 42x^5 - 210x^4.

To determine the intervals where the function f(x) is both decreasing and concave up, we need to find the values of x for which f'(x) < 0 (decreasing) and f''(x) > 0 (concave up).

Let's set f'(x) < 0 to find the decreasing intervals:
7x^6 - 42x^5 < 0.

Factoring out x^5, we get:
x^5(7x - 42) < 0.

Since x^5 is always positive, we only need to consider the sign of (7x - 42). To solve for x, we set (7x - 42) = 0 and find x = 6.

By using a sign chart or by selecting test points from each interval, we can determine that (7x - 42) < 0 when x < 6 and (7x - 42) > 0 when x > 6.

Now, let's set f''(x) > 0 to find the concave up intervals:
42x^5 - 210x^4 > 0.

Factoring out x^4, we have:
x^4(42x - 210) > 0.

Again, we only need to consider the sign of (42x - 210). Setting (42x - 210) = 0, we find x = 5.

By using a sign chart or test points, we can determine that (42x - 210) < 0 when x < 5 and (42x - 210) > 0 when x > 5.

Combining the intervals, we have:
-∞ < x < 5 and x > 6.

Therefore, the function f(x) = x^7 - 7x^6 is both decreasing and concave up on the intervals (-∞, 5) and (6, ∞).