Question

find a point c satisfying the conclusion of the mean value theorem for the given function and interval

y=x/(x+2) [1,4]

Answer 1

\[f \prime x=\frac{ -x }{ (x+2)^{2} }\]

Answer 2

\[\frac{ \frac{ 4 }{ 4+2 }-\frac{ 1 }{ 1+2 } }{4-1 } \]

Answer 3

\[\frac{ \frac{ 2 }{ 3 } -\frac{ 1 }{3 }}{3 }\]

Answer 4

\[\frac{ 1 }{ 3 }*\frac{ 1 }{ 3 }=\frac{ 1 }{ 9 }\]

Answer 5

\[\frac{ -x }{ (x+2)^{2} }=\frac{ 1 }{9 }\]

Answer 6

\[-9x=x ^{2}+4x+4\]

Answer 7

\[0=x ^{2}+13x+4\]

Answer 8

i know im supposed to factor now but i see no way of factoring

Answer 9

@Shadow_helper

Answer 10

sorry dude ... i got no idea... try @YanaSidlinskiy

Answer 11

@TheSmartOne

Answer 12

@brainzonly

Answer 13

\[f \prime(c)=\frac{ f(b)-f(a) }{ b-a }\]

Answer 14

@mathmale @Nnesha

Answer 15

\[f \prime(x)=\frac{ (x+2)*1-x*1 }{ (x+2)^2 }=\frac{ 2 }{ (x+2)^2 }\]
\[\frac{ 2 }{ (c+2)^2 }=\frac{ \frac{ 4 }{ (4+2) } -\frac{ 1 }{ (1+2) }}{ 4-1 }=\frac{ \frac{ 2 }{ 3 }-\frac{ 1 }{ 3 } }{ 3 }=\frac{ \frac{ 1 }{ 3 } }{ 3 }=\frac{ 1 }{ 9 }\]
\[(c+2)^2=18,c+2=\pm \sqrt{18}=\pm3\sqrt{2},c=-2 \pm3\sqrt{2}\]
now \[c=-2+3\sqrt{2}\in (1,4)\]

Answer 16

\[c=-2+4.24=2.24 \in (1,4) \]

Answer 17

1) Where did c come from
2) why is it equal to 18
3)what did you do?

Answer 18

read the statement of your question.

Answer 19

c is value of x.

Answer 20

ok but why 18 and what came after that because i see what you did but i do not know why

Answer 21

You differentiated incorrectly at tayfish

Answer 22

\[f \prime(c)=\frac{ 2 }{ (c+2)^2 }=\frac{ 1 }{ 9 }\]
cross multiply
\[(c+2)^2=9*2,c=2=\pm 3\sqrt{2}\]

Answer 23

at c the tangent to curve is parallel to the chord joining the points on the curve where x=1 andx= 4

Answer 24

correction above \[c=-2\pm 3\sqrt2 \]

Answer 25

ooohhhhhh oookkk thank you soo much!!!!

Answer 26

yw