Answer to Question #114642 in Calculus for ANJU JAYACHANDRAN

Question #114642
Find the intervals on which the function f defind by f(x)=x^4-8x^2+16 is concave upward or concave downward.
1
Expert's answer
2020-05-15T17:29:23-0400

Given

f(x)=x48x2+16f(x)=x^4-8x^2+16

Since

f(x)=4x316xf(x)=12x216f'(x) = 4x^3-16x\\ f''(x)= 12x^2-16

Solve

f(x)=0f''(x)=0

we get

x=233,    x=233x=\frac{2\sqrt{3}}{3},\ \ \ \ \:x=-\frac{2\sqrt{3}}{3}

Now , we check the sign of f(x)f''(x) , since

f(2)>0,   f(0)<0,  f(2)>0f''(-2)>0,\ \ \ f''(0)<0,\ \ f''(2)>0

Then f(x)f(x) is concave up on (,23/3)(23/3,)(-\infty , -2\sqrt{3}/3)\cup (2\sqrt{3}/3,\infty) and

concave down on (23/3,23/3)(-2\sqrt{3}/3,2\sqrt{3}/3)


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