The local minimum of function is an argument x for which the first derivative of function g(x) is equal to zero, so:
g'(x)=0
g'(x)=(x^4-5x^2+4)'=4x^3-10x=0
x(4x^2-10)=0
x=0 or 4x^2-10=0
4x^2-10=0 /4
x^2-10/4=0
x^2-5/2=0
[x-sqrt(5/2)][x+sqrt(5/2)]=0
Now we have to check wchich argument gives the minimum value from x=0, x=sqrt(5/2) and x=-sqrt(5/2).
g(0)=4
g(sqrt(5/2))=25/4-5*5/2+4=4-25/4=-9/4
g(-sqrt(5/2))=-9/4
The answer is sqrt(5/2) and -sqrt(5/2).
Answer:
64pi cm^2
i think
Step-by-step explanation:
Answer:
D
Step-by-step explanation:
Substitute the f(x) and g(x) into f/g
![\frac{\sqrt[3]{3x} }{5x+2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B3%5D%7B3x%7D%20%7D%7B5x%2B2%7D)
To find the domain,
Get the denominator of the fraction so 5x + 2 and equal it to 0 and find x
5x + 2 = 0
5x = -2
x = 
Therefore it's D
The answer is C.
Octagon is 135
And decagon is 144
So 144-135=11
Answer:
X=16
Step-by-step explanation:
1/4x-1=3
1/4x=3+1
1/4x=4
(4)1/4x=4(4)
x=16
