5 (b) iv (1 Viewer)

[mitch_07]

yeah i got something right lol my answer was 0.74

 

[Mushy18]

yes!!! 0.71....
but that's nowhere near 0.5...
????

 

[nit]

more like 0.5 is nowhere near 0.71

 

[acmilan]

for iv, said it was a root coz they intersect when y=x...??
ie. x = 1/(1 + x²)
which gets you to the x³ + x - 1 = 0

is that right..?

You need to find the intersection of f(x) and inverse of f(x)

A function and its inverse intersect at y = x

hence make f(x) = x to find where they intersect

1/(1+x²) = x
which gives x³ + x - 1 = 0

Then you need to find a value for x in this equation to find the x value of the intersection of the function and its inverse so you let f(x) = x³ + x - 1 and solve using Newtons method

 

[trickpat23]

Cheers everyone. Next time im walkin down the street and someone asks me that very question, in general conversation, i just say oh about 0.7. Thats what i love about maths, its so practical

 

[Heinz]

Cheers everyone. Next time im walkin down the street and someone asks me that very question, in general conversation, i just say oh about 0.7. Thats what i love about maths, its so practical

Learn something new every day

 

[gordo]

especially circle geometry

 

[egnaro]

i think you suppose to use the f(x)=1/(1+x^2)

not x^3+x-1=0

nope, you use x^3 + x - 1 since it is already stated that alpha is a root of the eqn and in the next bit it states that given 0.5 is a approximation for alpha. Therefor you use the eqn in part iv) not the original one.

and also, if you wrote that assuming x^3+x-1 is the eqn but they actually wanted you to use 1/(1+x^2) they would give you the mark