Question

What is the local minimum value of the function?

g(x)=x⁴-5x²+4

Answer 1

you need to differentiate the fuction and set it equal to zero. \[g \prime(x)=4x ^{3}-10x\]\[0=4x ^{3}-10x\]\[x=\sqrt{5/2}\] so \[g(\sqrt{5/2})\]should give you the answer

Answer 2

-9/4 is the answer in this case

Answer 3

differentiate and equate to 0:
4x^3 - 10x = 0
2x(2x^2 - 5)=0
x = 0 or +- sqrt (5/2)

so there re turning points at the x values above

find second derivative = 12x^2 - 10

now plug in the above values of x into second derivative - the positive value will give you the local minimum

Answer 4

yup zeros are sqrt(5/2), -sqrt(5/2) and 0

Answer 5

and then if I round to the nearest hundredth

Answer 6

-2.25

Answer 7

first you have to factorise it like this
lets say x^2=a
\[a^2-5a+4=0\]
\[(a-4)(a-1)=0\]
now you can write x^2 to places where you see a
\[(x-2)(x+2)(x-1)(x+1)=0\]
this is our equation
than try to make this minimum by giving values

Answer 8

local minimums are at x= sqrt(5/2), x= - sqrt(5/2)

Answer 9

yeah but it is an even function so both of them will give the same value thats why i didnt specify :)

Answer 10

yea - no probs