Answer: Choice D)
(-1.5, -1) and (0, 1)
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Explanation:
Exponents can be a bit clunky if you have too many of them, and if they're nested like this. Writing something like e^(x^2) may seem confusing if you aren't careful. I'm going to use a different notation approach. I'll use "exp" notation instead.
So instead of writing something like e^(x^2), I'll write exp(x^2).
The given derivative is
f ' (x) = exp(x^4-2x^2+1) - 2
and this only applies when -1.5 < x < 1.5
Apply the derivative to both sides and we'll find the second derivative
f ' (x) = exp(x^4-2x^2+1) - 2
f '' (x) = d/dx[ exp(x^4-2x^2+1) - 2 ]
f '' (x) = exp(x^4-2x^2+1)*d/dx[ x^4-2x^2+1 ]
f '' (x) = exp(x^4-2x^2+1)*(4x^3-4x)
f '' (x) = (4x^3-4x)*exp(x^4-2x^2+1)
From here, we need to find the roots of f '' (x).
Set f '' (x) equal to zero and solve to get...
f '' (x) = 0
(4x^3-4x)*exp(x^4-2x^2+1) = 0
4x^3-4x = 0 ..... or .... exp(x^4-2x^2+1) = 0
4x(x^2-1) = 0
4x(x+1)(x-1) = 0
4x = 0 or x+1 = 0 or x-1 = 0
x = 0 or x = -1 or x = 1
Those are the three roots. We ignore the equation exp(x^4-2x^2+1) = 0 because it doesn't have any real number solutions.
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The three roots of x = 0 or x = -1 or x = 1 represent possible locations of points of inflection (POI). Recall that a POI is where the function changes concavity. To determine if we have a POI or not, we'll need to a sign test.
Draw out a number line. Plot -1, 0, and 1 in that order on it. Pick something to the left of -1 but larger than -1.5, lets say we pick x = -1.2. Plugging this into the second derivative function leads to...
f '' (x) = (4x^3-4x)*exp(x^4-2x^2+1)
f '' (-1.2) = (4(-1.2)^3-4(-1.2))*exp((-1.2)^4-2(-1.2)^2+1)
f '' (-1.2) = -2.563
That value is approximate. The actual value itself doesn't matter. What does matter is the sign of the result. The negative second derivative value tells us we have a concave down region. So we just found that f(x) is concave down for the interval -1.5 < x < -1, which converts to the interval notation (-1.5, -1)
Repeat the process for something between x = -1 and x = 0. I'll pick x = -0.5 and it leads to f '' (-0.5) = 2.63 approximately. The positive result tells us that we have a concave up region. Therefore, -1 < x < 0 is not part of the answer we're after.
Repeat for something between x = 0 and x = 1. I'll pick x = 0.5 and it produces f '' (0.5) = -2.63 approximately. So the region 0 < x < 1 is also concave down. Meaning that the interval notation (0,1) is also part of the answer.
So far we have the interval notation of (-1.5, -1) and (0,1) as part of our solution set.
Lastly, we need to check something to the right of x = 1, but smaller than 1.5; let's go for x = 1.2
You should find that f '' (1.2) = 2.563 which allows us to rule out the region on the interval 1 < x < 1.5
Overall, the final answer is (-1.5, -1) and (0, 1)
