Question

Let f(x)=x+sqrt(1−x)
Find the local maximum and minimum values of f using both the first and second derivative tests.

Answer 1

I'm having a problem trying to solve the first derivative.
I get the the point \[-1 equals \frac{ 1 }{ 2\sqrt{1-x} }\]
but i don't get how to make it equal 3/4

Answer 2

please please help

Answer 3

\[f(x) = x + \sqrt{1 - x}\]\[f'(x) = 1 - \frac{1}{2\sqrt{1 - x}}\]

Answer 4

f'(x) = 0, then solve for x

Answer 5

\[1-\frac{ 1 }{ 2\sqrt{1-x} }=0\]
\[\frac{ 1 }{ 2\sqrt{1-x} }=1\]
\[2\sqrt{1-x}=1\]
\[\sqrt{1-x}=\frac{ 1 }{ 2 }\]
can you continue??

Answer 6

no i got there i don't know what to do next

Answer 7

You may have made a mistake when you took the derivative of
\[ \sqrt{1−x}= (1-x)^{\frac{1}{2}} \]
you should get
\[ \frac{1}{2} (1-x)^{(\frac{1}{2}-1)} \frac{d}{dx}(-x)\]
using the chain rule
that simplifies to

\[ \frac{1}{2} (1-x)^{-\frac{1}{2}} \cdot -1\]
or
\[ \frac{-1}{2\sqrt{1-x}} \]

Answer 8

so you should solve
\[ 1 - \frac{1}{2\sqrt{1-x}} = 0\]

Answer 9

yes i said that :)

Answer 10

to solve, multiply by 2 sort(1-x) to get rid of the denominator
move the -1 to the other side
square both sides to get rid of the square root.

Answer 11

thanks i got the right answer

Answer 12

welcome :)