Respuesta :

Answer:

Zero of the function;

f(x) = 0 at x = -1.333

Discontinuity;

x = 1

Step-by-step explanation:

To determine the zero of the function;

[tex]f(x)=\frac{3x^{2}+x-4 }{x-1}[/tex]

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

Ver imagen Hulkk
lublana

Answer:

The discontinuity of the given function f(x)=[tex]\frac{3x^2+x-4}{x-1}[/tex] at x=1

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

Step-by-step explanation:

Given function

f(x)=[tex]\frac{3x^2+x-4}{x-1}[/tex]

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

f(x)=[tex]\frac{3(1)^2+1-4}{1-1}[/tex]

f(x)=[tex]\frac{0}{0}[/tex] ( 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

[tex]\frac{3x^2+x-4}{x-1}[/tex]=0

By splitting middle term of numerator we get

[tex]\frac{3x^2+4x-3x-4}{x-1}[/tex]=0

By factorization we get

[tex]\frac{(3x+4)(x-1)}{x-1}[/tex]=0

By simplification we get

[tex]3x+4[/tex]=0

[tex]x=-\frac{4}{3}[/tex]

x=-1.33

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