CALC HOMEWORK!! PLEASE HELP!! Determine where the function is concave upward and where it is concave downward. f(x) = 3x^4 - 6x^3 +x - 9?
- anonymousLv 73 years ago
f(x) = 3x⁴ - 6x³ + x - 9
f '(x) = 12x³ - 18x² + 1
f ' '(x) = 36x² - 36x
The graph of f ' '(x) is an upwards-opening parabola (because the coefficient of the x² term is positive). When f ' '(x) > 0, the original function will be concave up. If you find the x-values where f ' '(x) = 0, you'll know that f ' '(x) > 0 outside of those points.
So here's f ' '(x) = 0 :
36x² - 36x = 0
x² - x = 0 ……………….. divided both sides by 36
x(x - 1) = 0 …………….. factored left side
By the principle of zero products, the solutions are x = 0 and x = 1.
f ' '(x) > 0 outside the interval (0, 1), and so the original function is concave up on the intervals (- ∞, 0) and (1, ∞).
The function is concave down when f ' '(x) < 0. In this case, it is concave down on (0, 1) .
Here's a graph of the function. I put in the points at x = 0 and x = 1 so that it's easier to see that the function is concave down on (0, 1).
There's also a hidden graph there of the second derivative function f ' '(x) = 36x² - 36x . If you click on the currently hollow circle next to that function in the left panel, it will appear. You can see from that graph where f ' '(x) is greater than and less than zero, which agrees with what was said above.
Note: I realized later something I said isn't always true, namely, the part about "If you find the x-values where f ' '(x) = 0, you'll know that f ' '(x) > 0 outside of those points." That works if you have an upwards-opening parabola with vertex below the x-axis for the second derivative . However, if the second derivative is a quadratic, it's also possible that there might be only one point where f ' '(x) = 0, or even no points where f ' '(x) = 0. Yet, the method is valid. If the second derivative is a quadratic, you can work with the graph of that quadratic to figure out where it is greater or less than zero.