  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

Step 1

Given:

Step 2

Obtain the second derivative in order to find the inflection point. help_outlineImage Transcriptionclose(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) fullscreen
Step 3

Concavity... help_outlineImage TranscriptioncloseBreak 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 fullscreen

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Trigonometry 