Question

Check my intervals of increase and decrease for function f(x)=2radical(x)-x.
I got
Dec:(-infinty,-1), (-1,1)
Inc: (1,2),(2,infinty)

Answer 1

what is radical(x) ??

Answer 2

Shouldn't the inc be (2,4) not (2,infinity)

Answer 3

\[f(x)=2\sqrt{x}-x \]
is the equation

Answer 4

whats your f'(x) ?

Answer 5

\[\frac{ 1 }{ \sqrt{x} }-1\]

Answer 6

correct
equating that to 0, you'll get x=1 as the only critical point, right ?

Answer 7

Yup

Answer 8

now, whats your f''(x)

Answer 9

\[\frac{ -1 }{2x ^{3/2} }\]

Answer 10

so tiny, its ^3/2

Answer 11

yeps,
f''(1) <0 , that means x=1 should be a point of maxima, right ?

Answer 12

yesss. thats what i got

Answer 13

Also, note that your function is continuous everywhere .

Answer 14

Hence, it must be inc in the interval (-inf,1) and dec in (1,inf)

at x=1, its neither inc not dec.

Answer 15

Any problems ? hope I didn;t make any mistake ?

Answer 16

oh am so sorry, I overlooked one thing

Answer 17

Im a little confused about finding the intervals, dont you pick a number in between 1 and something bigger like 2. so youd plug 1/2 into the first derivative to find it?

Answer 18

x<0 is not a part of the domain.

Answer 19

To find if its increasing or dec. sorry

Answer 20

see, f'(x) represents slope of the function.
f'(x) = 0 means we are finding that point where slope is 0, i.e. tangent is parallel to x axis.

we found x=1 is the only point there.
checking for f''(x), we find f''(1) < 0 which means rate of change of slope at x=1 is negative, or in other words, x=1 should be a point of maxima
the rough graph of the function must be something like this :


this signifies left of 1, graph should be increasing, to its right, it must be decreasing.

Answer 21

Am not following exactly what your query is ?

the answer should be increasing in [0,1) , and decreasing in (1, inf)

p

Answer 22

No that makes sense! I can look at the graph and see that. But i cant find it using the critical number if that makes sense.

Answer 23

you are trying to say, that you find f(1) , then you take any other point, say 2, and find f(2) ,
if f(1) > f(2) , function is deceasing to the right of 1 ,
same wahy f(1/2) < f(1) , so func is inc to the left of 1 ??

still not unable to understand your query ?