The function
f(x)=−2x^3+6x^2+48x−6
is increasing on the interval ( ).
It is decreasing on the interval ( −∞, ) and the interval ( , ∞ ).
The function has a local maximum at ( )
you know that f is increasing where f' > 0
f' = -6x^2+12x+48 = -6(x^2-2x-8) = -6(x-4)(x+2)
So, now you know where f'=0
where is f' > 0 and f' < 0 ?
To determine when a function is increasing or decreasing, we need to find the intervals where the derivative of the function is positive (increasing) or negative (decreasing). The derivative of the function can tell us about the rate of change of the function at different points.
Let's begin by finding the derivative of the given function, f(x):
f'(x) = d/dx (-2x^3 + 6x^2 + 48x - 6)
= -6x^2 + 12x + 48
To find when f(x) is increasing, we need to find the intervals where f'(x) > 0. Solving the inequality -6x^2 + 12x + 48 > 0 will give us the interval where f(x) is increasing.
-6x^2 + 12x + 48 > 0
Dividing by -6 (which reverses the inequality):
x^2 - 2x - 8 < 0
Now we can factor the quadratic equation:
(x - 4)(x + 2) < 0
To determine the sign of this expression, we need to consider the signs of the factors, (x - 4) and (x + 2), and their combinations. We can create a sign chart to help us visualize this:
| - | -2 | 4 | +
(x - 4) | - | - | 0 | +
(x + 2) | 0 | + | + | +
Product | + | - | 0 | +
From the chart, we see that the product of (x - 4) and (x + 2) is negative when -2 < x < 4. Therefore, the function f(x) is increasing on the interval (-2, 4).
To find when f(x) is decreasing, we need to find the intervals where f'(x) < 0. So, we solve the inequality -6x^2 + 12x + 48 < 0:
-6x^2 + 12x + 48 < 0
Dividing by -6:
x^2 - 2x - 8 > 0
Factoring:
(x - 4)(x + 2) > 0
From the sign chart we created earlier, we can see that the product of (x - 4) and (x + 2) is positive for x < -2 and x > 4. Therefore, the function f(x) is decreasing on the intervals (-∞, -2) and (4, ∞).
To find a local maximum, we need to identify points where the function changes from increasing to decreasing. One way to find these points is to set the derivative f'(x) equal to zero and solve for x:
-6x^2 + 12x + 48 = 0
Dividing by -6:
x^2 - 2x - 8 = 0
Factoring:
(x - 4)(x + 2) = 0
Setting each factor equal to zero:
x - 4 = 0, x + 2 = 0
x = 4, x = -2
The values x = 4 and x = -2 are the critical points. To determine if these are local maxima, we need to analyze the behavior of f'(x) around these points. We can use the second derivative test to determine if the critical points are local maxima or minima. However, since we are only interested in local maxima, we can evaluate the value of f''(x) at each critical point:
f''(x) = d^2/dx^2 (-2x^3 + 6x^2 + 48x - 6)
= -12x + 12
For x = 4:
f''(4) = -12(4) + 12
= -48 + 12
= -36
Since f''(4) < 0, the function does not have a local maximum at x = 4.
For x = -2:
f''(-2) = -12(-2) + 12
= 24 + 12
= 36
Since f''(-2) > 0, the function has a local maximum at x = -2.
Therefore, the function f(x) = -2x^3 + 6x^2 + 48x - 6 is increasing on the interval (-2, 4), decreasing on the intervals (-∞, -2) and (4, ∞), and has a local maximum at (-2, f(-2)).