The numerator is always positive, so the denominator determines the sign of f"(x). Thus, f"(x) is positive, and so f(x) is concave ---Select--- ♥ if x² > conclude that f"(x) is negative, and so f(x) is concave --Select--- ♥ h..... .... which means x > or x < We can also if < x < Therefore the interval where the function is concave up is as follows. (Enter your answer using interval notation.) The interval where the function is concave down is as follows. (Enter your answer using interval notation.)

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Please  explain step 2 clearly, this is my 3rd time posting the same question due to people not highlighting where the answer is..

 

Find the interval where the function is concave up. Find the interval where the function is concave down.
Step 1
-8x
Since f'(x) =
then
(x2 -
4)2'
(x² – 4)²(-8|
|-8 ) - (-8»)[2C«² – 4»(2- 2 *)
f"(x)
(x2.
4)
8 3
3
+
4
4
(x2 – 4)
Step 2
The numerator is always positive, so the denominator determines the sign of f"(x). Thus, f"(x) is positive,
and so f(x) is concave ---Select--- ♥ if x² >
conclude that f"(x) is negative, and so f(x) is concave ---Select--- ♥
which means x >
or x <
. We can also
if
< x <
Therefore the interval where the function is concave up is as follows. (Enter your answer using interval
notation.)
The interval where the function is concave down is as follows. (Enter your answer using interval notation.)
Transcribed Image Text:Find the interval where the function is concave up. Find the interval where the function is concave down. Step 1 -8x Since f'(x) = then (x2 - 4)2' (x² – 4)²(-8| |-8 ) - (-8»)[2C«² – 4»(2- 2 *) f"(x) (x2. 4) 8 3 3 + 4 4 (x2 – 4) Step 2 The numerator is always positive, so the denominator determines the sign of f"(x). Thus, f"(x) is positive, and so f(x) is concave ---Select--- ♥ if x² > conclude that f"(x) is negative, and so f(x) is concave ---Select--- ♥ which means x > or x < . We can also if < x < Therefore the interval where the function is concave up is as follows. (Enter your answer using interval notation.) The interval where the function is concave down is as follows. (Enter your answer using interval notation.)
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