Question

WILL GIVE BRAINLIEST: What is the local minimum value of the function g(x)=x^4-5x^2+4? (Round answer to the nearest hundredth)

Answer 1

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