Question

What is the discontinuity and zero of the function f(x) = the quantity of 3 x squared plus x minus 4, all over x minus 1 ?

Answer 1

The discontinuity of the given function f(x)=
`(3x^2+x-4)/(x-1)` at x=1

and zero of function f(x) is x=-1.33

Explanation:

Given function

f(x)=
`(3x^2+x-4)/(x-1)`

When we put x=1 then we get an indeterminate form

f(x)=
`(3(1)^2+1-4)/(1-1)`

f(x)=
`(0)/(0)` ( indeterminant form)

Therefore , the function is discontinuous at x=1 .Hence, the discontinuity at x=1.

Now, we find zero of given function by putting f(x)=0

f(x)=0

`(3x^2+x-4)/(x-1)`=0

By splitting middle term of numerator we get

`(3x^2+4x-3x-4)/(x-1)`=0

By factorization we get

`((3x+4)(x-1))/(x-1)`=0

By simplification we get

`3x+4`=0

`x=-(4)/(3)`

x=-1.33

Answer 2

Zero of the function;

f(x) = 0 at x = -1.333

Discontinuity;

x = 1

Explanation:

To determine the zero of the function;

`f(x)=(3x^(2)+x-4 )/(x-1)`

we simply graph the function as shown in the attachment below.

From the attachment, we can deduce that f(x) = 0 at x = -1.333

To determine the discontinuity or the undefined points of the function, we simply equate the denominator to 0 and solve for x;

x - 1 =0

x = 1

The function is therefore not defined at the point where x =1