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Consider the function f(x) = x^3-12x+8?
a) Find the interval where the graph is concave upward. (Enter your answer using interval notation.)
b) Find the interval where the graph is concave downward. (Enter your answer using interval notation.)
1 Answer
- llafferLv 71 year ago
The curve is concave upwards if the first derivative is positive:
f(x) = x³ - 12x + 8
f'(x) = 3x² - 12
3x² - 12 > 0
3x² > 12
x² > 4
If this was an equation, we'd get ±2 on the right side. But since this is an inequality, we have to flip the sign for the minus side, so:
x > 2 and x < -2
So the curve is concave upward:
(-∞, -2) U (2, ∞)
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It's then downward when the first derivative is negative.
3x² - 12 < 0
3x² < 12
x² < 4
x < 2 and x > -2
or:
-2 < x < 2
or:
(-2, 2)