Quick Question: Graphs (1 Viewer)

[Jonneeh]

How would you graph: (x+1)^4/x^4+1

Cant seem to get my head around it...

 

[Omar-Comin]

on what domain?

 

[Jonneeh]

on what domain?

its not indicated

 

[Omar-Comin]

you could differentiate, then find where x is max/min.
then find y intercept, some other random points, etc, and get a general shape

 

[Trebla]

Use division of functions y = (x + 1)⁴ and y = x⁴ + 1

 

[Jonneeh]

Use division of functions y = (x + 1)⁴ and y = x⁴ + 1

Could you elaborate abit please =)

 

[SeCKSiiMiNh]

my mx2 teacher always said that if you ever get stuck with graphing in an exam, you could always plot the points.

though he was kinda stupid, so noone took him seriously

 

[Omar-Comin]

Could you elaborate abit please =)

same as addition of functions but you divide..i.e draw the two curves then divide y1 by y2

 

[khorne]

Could you elaborate abit please =)

As above, it's under the technique of adding/manipulating ordinates. Generally, you should refrain from doing so, as it is considered a lesser technique, but in his case it is pretty straight forward, due to the numerator and denominator.

Graph both the numerator and denominator as separate functions on a single plan, then, at every x co-ordinate, divide the first functions y co-ordinate with the second functions co-ordinate to get the y co-ordinate of those functions combined.

The problem with this technique is that the graphs are rarely very accurate, as you are only doing them by increments.

 

[Trebla]

Could you elaborate abit please =)

Sketch y = (x + 1)⁴ and y = x⁴ + 1 on the same set of axes

- Notice that both curves are non-negative so the division of their functions is also non-negative.
- Both curves intersect at (0, 1), which means upon division the functions give a y-value of 1
- The x-intercepts of the numerator are always conserved (since the numerator = 0 causes the whole division of functions to be 0), hence (-1, 0) is a point and since the curve must be non-negative, by inspection this x-intercept must be the minimum point
- As x approaches positive or negative infinity, y approaches 1 since both numerator and denominator are quartic polynomials with positive leading coefficients
- Since (x + 1)⁴ > x⁴ + 1 for x > 0 (can see this in the sketch of both original functions) this implies (x + 1)⁴ / (x⁴ + 1) > 1 for x > 0, hence as x approaches infinity, the curve approaches y = 1 from above
- Now (x + 1)⁴ < x⁴ + 1 for x < 0 (can see this in the sketch of both original functions) this implies (x + 1)⁴ / (x⁴ + 1) < 1 for x < 0, hence as x approaches negative infinity, the curve approaches y = 1 from below

From this inspection work, you should now be able deduce the shape of the graph.