Question

Could anyone help me with this please?

Answer 1

Please see below.

Explanation:

Let us see what happens when `x` approaches `1` from left

Let us say `x=1-h` and then find `lim_(h->0)5cos(pi(1-h))`

It is evident that as `h->0`, `f(x)->-5`

Let us also see when `x` approaches `1` from right

Let us say `x=1+h` and then find the limit when `h->0`

Before that observe that

`(3x^2-x-2)/(1-x)=((3x+2)(x-1))/(1-x)=-(3x+2)`

Hence as `h->0` from right,

`lim_(x->1^+)=(3x^2-x-2)/(1-x)=-(3+2)=-5`

However, `f(x)` is not defined for `x=1`

and if we define `f(x)=-5` for `x=1`, then the function is continuous

i.e. `f(x)=[(cospix" if "x<1),(-5" if "x=1),((3x^2-x-2)/(1-x)" if "x>1)]`

Answer 2

See below.

Explanation:

`f(x)=[(5cos(pix), x<1),((3x^2-x-2)/(1-x), x >1)]`

Factor `3x^2-x-2`

`3x(x-1)+2(x-1)`

`(x-1)[3x+2]=(3x+2)(x-1)`

`((3x+2)(x-1))/(1-x)`

`((3x+2)(x-1))/(-(-1+x))=>((3x+2)(x-1))/(-(x-1)`

Removable discontinuity.

`((3x+2)cancel((x-1)))/(-cancel((x-1)))=(3x+2)/(-1)=-3x-2`

`f(x)=[(5cos(pix), x<1),(-3x-2, x >=1)]`