calculas help!!! (1 Viewer)

[da_butterfree]

hey guys.. i've been facing problems with graphing these curves...

it say find the turning points and plot the graphs

f(x)= x^7

and F(x) = x^5

they are supposed to look like the tan graph.. both at teh origin.. but i cant work it out.. everytime i do it, i just get a minimum...

can ne1 help??

 

[abdooooo!!!]

whats a turning point?

the graph kinda stops at origin... i believe its called a point of horizontal inflexion or something.

both should look the same but x^5 has more width relative to the x-axis since it has smaller power to compensate for.

 

[abdooooo!!!]

just draw 2 prabolas at origin one is under the x-axis and the other above.

then rub out the left top quadrant and bottom right quadrant

 

[da_butterfree]

umm.. ther are both seperate questions.. i do understand what the graph looks like.. but the working out it tricky..
sum 1.
f(x) = x^5
f'(x) = 5x^4 when f'(x)=0, x=0 and y=0.
then f''(x) = 20x^4 so when f''(x) = 0, x=0 hence it can be a point of horizontal inflexion. but when we check for the inflexion it proves to be a minmum.. how is that??

 

[abdooooo!!!]

its not possible... i don't know what you did but here is my reasoning:

x^3 is the second deravitive of x^5

substitute 1 and -1

(-1)^3 = -1

(1)^3 = 1

thus change in concavity

on the left side of 0 the graph is concave down and on the right side it is concave up. so its not a minimum or a maximum.

 

[da_butterfree]

ohh!! but how did u get f''(x) = x^3?? when f(x) is x^5??
and neways there is a change in concavity.!

 

[CM_Tutor]

Both y = x⁵ and y = x⁷ have horizontal points of inflection at the origin, so both look like the standard y = x³ cubic, but flatter through the inflection and steeper after that. Neither is particularly similar to y = tan x beyond general shape, as tan x is trapped between vertical asymptotes, and is diagonal through the origin (m = 1), whereas these have no vertical asymptotes and are horizontal through the origin.

 

[fatmuscle]

if anyone wants a graphic program, PM me

 

[smegthehead]

a good free graph program is availible here

 

[Grey Council]

hurm, all equations that are simply:
x^n, where n is an odd number, look like the x^3 graph, but the higher the number n is, the flatter at the origin.

if n is even, then it looks like x^2, but flatter at the origin.

try it out in the graphing program posted above.

 

[Collin]

x^3 is the second deravitive of x^5

I think you forgot about the co-efficients.

 

[kpq_sniper017]

For f(x)=x^5

f'(x)=5x^4
f"(x)=20x^3

f'(x)=0 when x=0

so there's a stationary point at x=0 y|x=0 : 0
so the stat. pt is at the origin

test values of x either side of x=0
f"(-0.1) : <0
f"(0.1)" >0

so therefore, concavity changes
i.e. theres a horizontal point of inflexion at (0,0)

same pretty much applies for x^7 (except the graph will looking slightly different).