Question

HELP! THIS THE THIRD TIME I HAVE BUMBED THIS. Help appreciated! What is the absolute min/max of

x(sqrt(x^2+16))
I got this derivative sqrt(x^2+16)+(x^2)/(sqrt(x+16), which you set to zero to find the min/max. I cant seem to find them!

Answer 1

is this your function:
\[f(x)=x \sqrt{x ^{2}+16}\]

Answer 2

yes!

Answer 3

\[f \prime(x)=\sqrt{x ^{2}+16}+\frac{ x }{ 2\sqrt{x ^{2}+16} }\]=0

Answer 4

take l.c.m
\[\frac{ 2(x ^{2}+16)+x }{ 2\sqrt{x ^{2}+16} }=0\]

Answer 5

I have and I cant seem to find the min/ or max I keep getting + -2sqrt2

Answer 6

\[2x^{2}+x+32=0\]
solve for x

Answer 7

okay. you want me to solve this further.

Answer 8

yes please!

Answer 9

is there any range in which this function is defined?

Answer 10

(-5,7)

Answer 11

can you tell me how you got \[\pm2\sqrt{2}\]

Answer 12

honestly, it was wolframalpha since, I cant seem to get through my arithmetic fully

Answer 13

okay, just find out the roots of that equation
and plug in all four values (2 values you will find out and 2 values are -5 and 7)
in \[f \prime(x)\]
and check on which value the function is minimum and maximum.

Answer 14

the value at which the function is minimum is absolute minima
and the value at which the function is maximum is absolute maxima.
hope this will help.

Answer 15

yes, this is what im asking, I cant seem to find the roots

Answer 16

sorry., you have to plug in all 4 values in f(x) not in \[f \prime(x)\]

Answer 17

okay. use this formula
\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]

Answer 18

here the roots of your equation are imaginary.So you have to find the critical points of your function where it becomes zero.

Answer 19

ok so do I just se the equation to zero? to f prime?

Answer 20

sorry I cant seem to find it thank u tho.

Answer 21

can you help here... @hartnn

Answer 22

@neer2890 i think you made an error when you found the derrivative. you forgot to "do the inside" the x^2 which would be 2x

(x^2 + 16)^1/2 =1/2(x^2 +16)^-1/2*2x = x / (x^2 + 16)^1/2

Answer 23

oh... thanks for correcting me @triciaal

Answer 24

but the question still remains same. how to find the critical roots?

Answer 25

method :

equate the derivative to 0

if you don't find any real roots, then
use the end-points

plug in x = -5
and x =7

the smaller value will be miminum
larger value will be maximum

Answer 26

\[f \prime(x)=\frac{ 2x ^{2}+16 }{ \sqrt{x ^{2}+16} }=0\]

Answer 27

so the absolute minima=-32
and absolute maxima=56.4
is it?

Answer 28

its like this function,

minimum and maximum are just endpoints