Question

If g(x)=e^(2x)*f(x) and f(2)=-3, what is the equation of the normal line to the graph of g at x=2?

Answer 1

\[g(x)=e ^{2x}*f(x)\]

Answer 2

\[g'(x)=2e^{2x}*f(x)+ f ' (x)e^{2x}\]

Answer 3

when x=2 what do I put for f ' (2)

Answer 4

I might need to add a graph

Answer 5

give me a moment to upload

Answer 6

ok not sure that helps

Answer 7

what are your thoughts?@hartnn

Answer 8

that graph is of f'(x)
so f'(2) is just the value of that curve at x=2

Answer 9

f'(2) =4

Answer 10

sorry can't see your prime
did you put f ' (2)=4

Answer 11

yes

Answer 12

so the slope of the tangent line is -2e^4

Answer 13

so the slope of the normal line is 1/(2e^4)

Answer 14

correct

Answer 15

ok \[y+3=\frac{ 1 }{ 2e^4 }(x-2)\]

Answer 16

'y' value isn't -3

Answer 17

for g(x) its g(2)

Answer 18

oh so I have to calculate g(2), ok give me a moment

Answer 19

yes
normal line to graph of "g"
so find a point on 'g'
x=2, y =g(2)

Answer 20

g(2)=-3e^4

Answer 21

so the normal line is
\[y-2=\frac{ 1 }{ 2e^4 }(x+3e^4)\]

Answer 22

how does that look?

Answer 23

you interchanged x1 and y1

Answer 24

y- (-3e^4) = m (x-2)

Answer 25

yes, I see that now....

Answer 26

\[y+3e^4=\frac{ 1 }{ 2e^4 }(x-2)\]

Answer 27

is that better?

Answer 28

weird answer
but out calculations seem to be correct...

Answer 29

Can I ask you 4 more questions about that given f ' graph

Answer 30

sure

Answer 31

yes I have many weird answer problems

Answer 32

1. On the interval [0,8], are they any values where f(x) is not differentiable? Give a reason for your answer.

Answer 33

I want to say at x=2

Answer 34

not sure why

Answer 35

http://math.mit.edu/classes/18.013A/HTML/chapter06/section03.html

Answer 36

basically you cannot have a jump for a differentiable function

Answer 37

I guess I can use the justification of a limit

Answer 38

in other words, the slope should not change from + to - or - to + suddenly.

Answer 39

ok I think I can justify it with the jump description or the limit definition.

Answer 40

Next question, On what interval(s) is f " >0 f "(x)<0? Give reasons for your answers

Answer 41

f "(x)>0 means f " (x) is positive correct? or that f (x) is concave up

Answer 42

when f'(x) is increasing, f''(x) will be positive

Answer 43

so f " (x) >0 on (0,2) and (6,8)

Answer 44

yep

Answer 45

and f " (x) < 0 on (4, -2)

Answer 46

ok now I need to find two values
f " (4) and f " (8)
I don't know how to do this

Answer 47

f " (x) < 0 on (2,5)

and
f''(x) =0 on (5,6)

Answer 48

so am I just stating the slopes on those intervals

Answer 49

note that f " (x) < 0 on (4, -2) was incorrect.

i corrected it to
f " (x) < 0 on (2,5)

Answer 50

can you find the slope of line, given 2 points ?

Answer 51

yes f " (4) = -2

Answer 52

correct

Answer 53

\[\frac{ f(5)-f(2) }{ 5-2 }=\frac{-2-4 }{ 3 }=\frac{ -6 }{ 3 }=-2\]

Answer 54

what about f " (8)?

Answer 55

I guess I should have written f ' (5) - f ' (2) for the numerator on top

Answer 56

6,-2
8,2

Answer 57

ok I can use those two points

Answer 58

yes, thats correect
f'(5) and f'(2)

Answer 59

ok f " (8)=2

Answer 60

yes

Answer 61

Thanks you made it seem easy...

Answer 62

it is easy :)
welcome ^_^