Question

Consider the function f(x)=x^(2)e^(x)-2xe^(x)-3e^(x).
(A) Find the equation of the normal line to f where x=0.
(B) Find the value(s) of x for which f has a horizontal tangent.

Answer 1

can you find f' ?

Answer 2

Yeah would it be 2xe^(x)-2e^(x)-e^(x)

Answer 3

f(x)=x^(2)e^(x)-2xe^(x)-3e^(x).
f'(x)=e^x(x^2-5)
for horizontal tangents
f'(x)=e^x(x^2-5)=0 ---> x1=sqrt5=2.236 x2=-sqrt5=-2.236 plug them to orig eq to find your y

Answer 4

your normal line would be perpendicular to these tangent line

Answer 5

any question?

Answer 6

When plugging in the x-values what do I do with the e's.

Answer 7

plug in also the x values to e

Answer 8

Do I do both x-values at once or separately?

Answer 9

use your calculator

Answer 10

f(x)=sqrt5^(2)e^(sqrt5)-2sqrt5e^(sqrt5)-3e^(sqrt5)
this is also
f(x)=e^x (x^2-2x-3) plug in the x here

Answer 11

f(x)=e^x (x^2-2x-3)
f(sqrt5)=e^sqrt5 ((sqrt5)^2-2sqrt5-3)=-23
f(sqrt-5)=e^sqrt-5 ((sqrt-5)^2-2sqrt-5-3)=0.692
P(x,y)=P(sqrt5,-23)., Q(x,y)=Q(-sqrt5,0.692) these are your horizontal tangents

Answer 12

next is to find the normal line of f(x) at x=0

Answer 13

Yay I got the same answers as you :)

Answer 14

ok did you find the normal line eq?

Answer 15

very good

Answer 16

No what do I do to get that?

Answer 17

the normal line to f(x) will have a perpendicular to that line meaning the slope has a formula m=-1/m (slope for perpendicular line)

Answer 18

ist get your alope at x=0

Answer 19

slope m=f'(x)=e^x(x^2-5) now at x=0
m=e^0(0^2-5)=-5
now the line perpendicular to it will have a slope of
m=-1/m
m=-1/-5=1/5
also you y value at x=0 is y=f(0)=e^0 (0-2(0)-3)
y=-3
now you have (0,-3)
your equation normal to the line is
y-y1=m(x-x1)
y-(-3)=1/5 (x-0)
y+3=1/5 x
y=1/5 x -3 Normal line to f(x) ans

Answer 20

did you get the idea? any question,

Answer 21

Yeah thanks you helped immensely.Now I can do the rest of my homework problems.

Answer 22

very good,,good luck now..have fun ..lol :D

Answer 23

thanks :)

Answer 24

yw