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f(x)=−0.5⋅x^4+3x^2 what are the interverls for concavity

Question

 f(x)=−0.5⋅x^4+3x^2 what are the interverls for concavity 

 

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Step 1

Given:

The function is f(x)= -0.5x4+3x2
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The function is f(x)= -0.5x4+3x2

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Step 2

Obtain the second derivative in order to find the inflection point.

(x)0.5(4x)+3(2x) -2x +6x
f (x)2 (3x2)+6(1)- -6x +6
Set f" (x)0 and obtain the value ofx as 1 and-1
The value of the fbecomes
f(1)-0.5(1)3(t1)
=-0.53
= 2.5
Thus, the inflection points are (1,2.5) and (-1,2.5)
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(x)0.5(4x)+3(2x) -2x +6x f (x)2 (3x2)+6(1)- -6x +6 Set f" (x)0 and obtain the value ofx as 1 and-1 The value of the fbecomes f(1)-0.5(1)3(t1) =-0.53 = 2.5 Thus, the inflection points are (1,2.5) and (-1,2.5)

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Step 3

Concavity...

Break the interval (, ) into sub intervals (,-1) (-1,1) and (1,o)
Choose one number from each interval and check for the value of second derivative
When x 2 the second derivative becomes f"(-2) -6(-2) +6 =-18; hence the
function is concave down in the interval (--1)
When x0, the second derivative becomes f"(0)= -6(0)' +6 = 6; hence the function
is concave up in the interval (-1,1)
When x 2, the second derivative becomes f"(2) =-6(2) +6 =-18; hence the
function is concave down in the interval (1,o
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Break the interval (, ) into sub intervals (,-1) (-1,1) and (1,o) Choose one number from each interval and check for the value of second derivative When x 2 the second derivative becomes f"(-2) -6(-2) +6 =-18; hence the function is concave down in the interval (--1) When x0, the second derivative becomes f"(0)= -6(0)' +6 = 6; hence the function is concave up in the interval (-1,1) When x 2, the second derivative becomes f"(2) =-6(2) +6 =-18; hence the function is concave down in the interval (1,o

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Trigonometry

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