Question

Suppose we don't have a formula for g(x) but we know that g(2)=-4 and g'(x)+sqrt(x^2+5) for all x.
(a) use linear approximation to estimate g(1.95) and g(2.05)
(b) Are your estimates in part (a) too large or too small? Explain.

Answer 1

my estimates were \[g(1.95)\approx-5.5 \]\[g(2.05)\approx-2.5\]

Answer 2

i just don't get part b. if i had the graph i would look at the concavity but i dont

Answer 3

@hartnn

Answer 4

Is g'(x)+sqrt(x^2+5)=0 ?

Answer 5

g'(x) a function. g'(x) is the derivative of g(x)

Answer 6

sorry it was a typo. g'(x)=sqrt(x^2+5)

Answer 7

take out value of g from g(2)=-4 which is -8. replace -8 in the other equation for g

Answer 8

where did you get -8

Answer 9

you said you know that g(2)=-4, didnt you? solve it and you get -8. replace that for g in other equations.

Answer 10

@muzzammil.raza

Answer 11

for (b), look at the sign of g''(x)
(the higher order terms of a taylor expansion)

Answer 12

sorry websight failed.

Answer 13

@phi i dont know what taylor expansion is. @ichiro what are you solving

Answer 14

\[g'\left( x \right)=\int\limits \sqrt{x ^{2}+5}*1dx\]
\[g \left( x \right)=\sqrt{x ^{2}+5}*x-\int\limits \frac{ 2x }{2\sqrt{x ^{2}+5} }*x dx\]
\[g \left( x \right)=x \sqrt{x ^{2}+5}-\int\limits \frac{ x ^{2}+5-5 }{\sqrt{x ^{2}+5} }dx\]

Answer 15

what is that curvy line thing

Answer 16

\[g \left( x \right)=x \sqrt{x ^{2}+5}-\int\limits \sqrt{x ^{2}+5}dx+5\int\limits \frac{ dx }{\sqrt{x ^{2}+5} }\]
\[g \left( x \right)=x \sqrt{x ^{2}+5}-g \left( x \right)+5 \int\limits \frac{ dx }{\sqrt{x ^{2}+5} }\]

Answer 17

\[2g \left( x \right)=x \sqrt{x ^{2}+5}+5 I\]

Answer 18

i don't understand any of this

Answer 19

\[I=\int\limits \frac{ dx}{ \sqrt{x ^{2}+5} }\]
Let me complete first.

Answer 20

\[Putx=\sqrt{5}\tan \theta \]
\[dx=\sqrt{5}\sec ^{2}\theta d \theta \]

Answer 21

\[I=\int\limits \frac{ \sqrt{5}\sec ^{2}\theta d \theta }{\sqrt{5\tan ^{2}\theta+5} }\]
\[I=\int\limits \frac{ \sqrt{5}\sec ^{2}\theta d \theta }{\sqrt{5}\sec \theta }=\int\limits \sec \theta d \theta \]
\[=\ln \left| \sec \theta+\tan \theta \right|+c\]

Answer 22

\[I=\ln \left| \frac{ \sqrt{x ^{2}+5} }{ \sqrt{5} }+\frac{ x }{\sqrt{5} } \right|+c\]

Answer 23

\[2g \left( x \right)=x \sqrt{x ^{2}+5}+5[\ln \left| \frac{ x+\sqrt{x ^{2}+5} }{\sqrt{5} } \right|+c\]

Answer 24

\[g \left( x \right)=\frac{ x \sqrt{x ^{2}+5} }{ 2 }+\frac{ 5 }{ 2 }\ln \left| \frac{ x+\sqrt{x ^{2}+5} }{\sqrt{5} } \right|+\frac{ c }{ 2 }\]
put x=2 and find c

Answer 25

now you can ask me.
I have used integrate by parts.

Answer 26

sorry i have not done by linear approximation.