f(x)=x2(x−24).
So f is decreasing (and f′ is negative) on the interval (0,a)
for a= ?
To determine the value of "a," we need to find the critical points of the function f(x)=x^2(x-24) and identify where the function is decreasing.
Step 1: Find the derivative of the function f(x):
f'(x) = 2x(x-24) + x^2
Step 2: Set the derivative equal to zero to find the critical points:
2x(x-24) + x^2 = 0
Step 3: Simplify and solve for x:
2x^2 - 48x + x^2 = 0
3x^2 - 48x = 0
3x(x - 16) = 0
Setting each factor equal to zero:
3x = 0 => x = 0
x - 16 = 0 => x = 16
So, the critical points of the function are x = 0 and x = 16.
Step 4: To determine where the function is decreasing, we need to look at the sign of the derivative f'(x) on the interval (0, a).
Using test intervals:
For x < 0:
Choose x = -1, substitute it into f'(x):
f'(-1) = 2(-1)(-1-24) + (-1)^2 = -50
Since f'(-1) is negative, the function f(x) is decreasing for x < 0.
For 0 < x < 16:
Choose x = 5, substitute it into f'(x):
f'(5) = 2(5)(5-24) + (5)^2 = -237
Since f'(5) is negative, the function f(x) is decreasing for 0 < x < 16.
For x > 16:
Choose x = 20, substitute it into f'(x):
f'(20) = 2(20)(20-24) + (20)^2 = 120
Since f'(20) is positive, the function f(x) is increasing for x > 16.
Therefore, the function f(x)=x^2(x-24) is decreasing (and f'(x) is negative) on the interval (0, 16). Hence, a=16.
To find the value of a, we need to determine the interval on which the function f(x) = x^2(x - 24) is decreasing.
To do this, we will analyze the sign of the derivative, f'(x), since a function is decreasing when its derivative is negative.
First, let's find the derivative of f(x) using the product rule:
f(x) = x^2(x - 24)
f'(x) = 2x(x - 24) + x^2(1)
= 2x^2 - 48x + x^2
= 3x^2 - 48x
Now, we need to find the critical points of f'(x) by setting it equal to zero and solving for x:
3x^2 - 48x = 0
x(3x - 48) = 0
From this equation, we have two solutions: x = 0 and 3x - 48 = 0, which gives x = 16.
Next, we need to test the intervals (-∞, 0), (0, 16), and (16, +∞) to determine where f'(x) is negative.
For the interval (-∞, 0):
Choose a test point, let's say x = -1:
f'(-1) = 3(-1)^2 - 48(-1) = 3 + 48 = 51 (positive)
Since f'(-1) > 0, the function is not decreasing on this interval.
For the interval (0, 16):
Choose a test point, let's say x = 1:
f'(1) = 3(1)^2 - 48(1) = 3 - 48 = -45 (negative)
Since f'(1) < 0, the function is decreasing on this interval.
For the interval (16, +∞):
Choose a test point, let's say x = 17:
f'(17) = 3(17)^2 - 48(17) = 867 - 816 = 51 (positive)
Since f'(17) > 0, the function is not decreasing on this interval.
Therefore, f(x) is decreasing (and f'(x) is negative) on the interval (0, 16).
Hence, the value of a is 16.