find the intervals on which the function is increasing or decreasing and find any relative minima or maxima.
f(x)=0.2x^3-0.2x^2-5x-4
f=1/5 (x+4)(x^2 - 5x - 5)
f' = .6x^2 - .4x - 5
f'=0 at x = 1/3 (1 ± 2√19)
f increases where f' > 0
To find the intervals on which the function is increasing or decreasing, we need to analyze the derivatives of the function. Let's start by finding the first derivative of the function:
f'(x) = 0.6x^2 - 0.4x - 5
To find the intervals of increase and decrease, we need to locate the critical points of the function. These are the points where the derivative is equal to zero or does not exist.
Setting f'(x) = 0:
0.6x^2 - 0.4x - 5 = 0
Now, we can solve this quadratic equation to find the x-values of the critical points. Using the quadratic formula, we have:
x = (-(-0.4) ± √((-0.4)^2 - 4(0.6)(-5))) / (2 * 0.6)
Simplifying further, we get:
x = (0.4 ± √(0.16 + 12)) / 1.2
x = (0.4 ± √12.16) / 1.2
x = (0.4 ± 3.485) / 1.2
x ≈ -2.904 or x ≈ 3.237
These are the x-values of our critical points. To determine the intervals of increase and decrease, we can choose test points within these intervals and evaluate the sign of the first derivative at those points.
Let's choose a test point less than -2.904, like x = -5:
f'(-5) = 0.6(-5)^2 - 0.4(-5) - 5
f'(-5) = 14.5
Since f'(-5) > 0, the derivative is positive in this interval, which means the function is increasing.
Next, let's choose a test point between -2.904 and 3.237, like x = 0:
f'(0) = 0.6(0)^2 - 0.4(0) - 5
f'(0) = -5
Since f'(0) < 0, the derivative is negative in this interval, which means the function is decreasing.
Finally, let's choose a test point greater than 3.237, like x = 5:
f'(5) = 0.6(5)^2 - 0.4(5) - 5
f'(5) = 14.5
Since f'(5) > 0, the derivative is positive in this interval, which means the function is increasing.
To summarize:
- The function is increasing on the interval (-∞, -2.904) and (3.237, ∞).
- The function is decreasing on the interval (-2.904, 3.237).
To find any relative minima or maxima, we need to locate the x-values where the function changes from increasing to decreasing or vice versa. Since we have found that the function is increasing on (-∞, -2.904) and (3.237, ∞), and decreasing on (-2.904, 3.237), we can conclude that there is a relative minimum at x ≈ -2.904 and a relative maximum at x ≈ 3.237.
To find the corresponding y-values, we can substitute these x-values back into the original function:
f(-2.904) ≈ -14.348
f(3.237) ≈ -12.443
Therefore, the relative minimum is approximately (-2.904, -14.348) and the relative maximum is approximately (3.237, -12.443).