Jacob C. answered • 07/09/21
Adaptive Math and Physics Tutor
f(x) is decreasing when the derivative, f'(x), is negative. We must compute the derivative, determine the points at which it is equal to 0, and test points on either side of these critical points to determine the intervals in which f(x) is decreasing.
The derivative is simply f'(x) = 56x3 + 39x2 - 2x and we are looking for all the points x such that f'(x) = 0. The first and most obvious point at which f'(x) = 0 is x = 0. Next, we will find the other two roots.
56x3 + 39x2 - 2x = 0
x(56x2 + 39x - 2) = 0
56x2 + 39x - 2 = 0
By the quadratic equation:
x = (-39 ± √(392 + 448))/112
x1 ≈ 0.048
x2 ≈ -0.744
Then, knowing that f'(x) = 0 at x ≈ -0.744, x = 0, and x ≈ 0.048, we can choose points on each side of these points and test whether f'(x) is positive or negative. Choose x = -1, x = -0.5, x = 0.02, and x = 1 as test points and compute f'(x).
f'(-1) = 56(-1)3 + 39(-1)2 - 2(-1) = -56 + 39 + 2 = -15
f'(-0.5) = 56(-0.5)3 + 39(-0.5)2 - 2(-0.5) = -7 + 9.75 + 1 = 3.75
f'(0.02) = 56(0.02)3 + 39(0.02)2 - 2(0.02) = 0.000448 + 0.0156 - 0.04 = -0.023952
f'(1) = 56(1)3 + 39(1)2 - 2(1) = 56 + 39 - 2 = 93
Since f'(-1) < 0, f(x) is decreasing on the interval (-∞, (-39 - √1969)/112). Since f'(-0.5) > 0, f(x) is increasing on the interval ((-39 - √1969)/112, 0). Since f'(0.02) < 0, f(x) is decreasing on the interval (0, (-39 + √1969)/112). Finally, since f'(1) > 0, f(x) is increasing on the interval ((-39 + √1969)/112, ∞).
Thus, f(x) is decreasing on the interval (-∞, (-39 - √1969)/112) ∪ (0, (-39 + √1969)/112).
