What is the area of a polygon with vertices of (-2,-4), (4,-4), (4, 4), and (-5, 4)?
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Answer:
Step-by-step explanation:
you can find where the first derivative is 0 to find the critical points
g'(x) = ( x^4 -5x^2 +4)' = 4x³-10x
g'(x) =0, make y =0 to find find where g'(x) is 0
4x³-10x =0 , factor 2x
2x(2x²-5)= 0 , each factor must be 0
2x= 0, so x= 0
2x²-5 =0, so x = ±√5/2
we now have 3 critical points -√5/2, 0, and √5/2
make intervals (-∞, -√5/2), (-√5/2, 0) , (0, √5/2) and (√5/2, +∞)
pick a point to test on each interval: -2, -1, 1 and 2 for example, and
calculate g'(x) = 4x³-10x at those points
for x= -2 we have 4(-2)³-10(-2) = -12 , negative number, decrease
for x= -1 we have 4(-1)³-10(-1) =6, positive number, increase
for x= 1 we have 4(1)³-10(1) = -6, negative number, decrease
for x= 2 we have 4(2)³-10(2) = 12, positive number, increase
we went from a decrease to an increase on intervals (-∞, -√5/2), (-√5/2, 0) so x= - √5/2 ≈ -1.58 is a minimum
we went from a decrease to an increase on intervals (0, √5/2), (√5/2, +∞) so
x= √5/2 ≈ 1.58 is a minimum as well
