Question

Find critical numbers of the function
g(x)=x√4-x

Answer 1

g(x)=x(4-x)^1/2
g'(x)=(4-x)^1/2+[1/2(4-x)^-1/2]
what do i do next

Answer 2

you mean critical point , right ?
so,
Take the derivative of function to find g'(x) and equate it to 0
g'(x) = 0
then the values of x are the critical poitns

Answer 3

the 2nd term of your derivative is incorrect, maybe check again ?

Answer 4

\[g(x)=x(4-x)^\frac{ 1 }{ 2 }\]
\[g'(x)=(4-x)^\frac{ 1 }{ 2 }+ \frac{ 1 }{ 2 }x(4-x)^-\frac{ 1 }{ 2 }\]

Answer 5

now thats correct :)

Answer 6

wait did you miss the negative sign for the 2nd term ?

Answer 7

\[g′(x)=(4−x)^\frac{ 1 }{ 2}+[\frac{ 1 }{ 2 } x(4−x)^-\frac{ 1 }{ 2 }]\]

Answer 8

because derivative of 4-x is -1

Answer 9

won't you get
\(g′(x)=(4−x)^\dfrac{ 1 }{ 2}-[\dfrac{ 1 }{ 2 } x(4−x)^{-\dfrac{ 1 }{ 2 }}]\)
??

Answer 10

then equating g'(x) = 0 gives
\(g′(x)=(4−x)^\dfrac{ 1 }{ 2}-[\dfrac{ 1 }{ 2 } x(4−x)^{-\dfrac{ 1 }{ 2 }}]=0 \\or, (4-x) = x/2\)

Answer 11

It is only a negative when you are doing the quotient rule. This is the product rule that we are using.

Answer 12

ys, and the chain rule too!
heard of it ?

Answer 13

Oh yea

Answer 14

d/dx (sqrt (4-x)) = 1/2sqrt (4-x) d/dx (4-x) = - 1/2 sqrt(4-x)

Answer 15

so can you continue now ?

Answer 16

yes/no/idk ?

Answer 17

@hartnn is correct, the second term will be negative due to the chain rule.