Question

CAN SOMEONE PLEASE HELP ME FIND THE ABSOLUTE MINIMUM AND ABSOLUTE MAXIMUM OF THIS EQUATION? IM SO CONFUSED

Answer 1

\[y=x \sqrt{25-x^2}\]

Answer 2

\[Let s=y ^{2}=x ^{2}\left( 25-x ^{2} \right)=25x ^{2}-x ^{4}\]
\[\frac{ ds }{ dx }=50x-4x ^{3}\]

Answer 3

so what do i do next?

Answer 4

\[\frac{ d ^{2}s }{dx ^{2} }=50-12x ^{2}\]

Answer 5

\[find stationary values by plugging \frac{ ds }{dx }=0,then for these values find y \]
you can check for maxima and minima by plugging in \[\frac{ d ^{2}s }{dx ^{2} },if \frac{ d ^{2} s}{dx ^{2} }>0,then relative minima,\]

Answer 6

\[if \frac{ d ^{2}s }{ dx ^{2} }<0,then relative maxima.\]

Answer 7

i have no idea what any of that means. i've never learned any of that.. we're supposed to do it on the graphing calculator

Answer 8

y exists only if
\[25-x ^{2}\ge 0,or 25\ge x ^{2},x ^{2}\le 25,\left| x \right|\le 5,-5 \le x \le 5\]

Answer 9

\[50x-4x ^{3}=0,2x \left( 25-2x ^{2} \right)=0,\]

Answer 10

Either 2x=0,gives x=0
\[or 25-2x ^{2}=0,x=\pm \frac{ 5 }{\sqrt{2} }\]

Answer 11

\[so x=-5,\frac{ -5 }{\sqrt{2} },0,\frac{ 5 }{\sqrt{2} },5\]

Answer 12

find y for these values and find absolute maximum and absolute minimum value.

Answer 13

how would i find y

Answer 14

\[put the values of x \in y=x \sqrt{25-x ^{2}}, one by one.\]

Answer 15

show me your work after solving for x=-5

Answer 16

so i would make x=0? y=0\[\sqrt{25-0^2}\]

Answer 17

\[when x=-5,y=-5\sqrt{25-25}=-5*0=0\]

Answer 18

put other 4 values

Answer 19

\[when x=\frac{ -5 }{\sqrt{2}} ,y=-\frac{ -5 }{\sqrt{2} }\sqrt{25-\frac{ 25 }{2 }}=\frac{ -5 }{ \sqrt{2} }*\frac{ 5 }{\sqrt{2} }=\frac{ -25 } { 2 }\]

Answer 20

similarly find y for x=0,x=5/sqrt2,x=5