# Determine The Left And Right Behavior Of The Graph Of The Polynomial Function f(x)=3+x-4x^4

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f(x) = 3+x-4x^4.

To determine the behaviour of the graph.

When x = 0 , y = 3. So the graph makes an intercept of 3 on y axis.

When y= 0, 3+x-4x^3 is satisfied for x= 1.

Therefore x= 1, is an x axis intercept.

Similarly there is one more intercept of x axis between 0 and -1.

Also f'(x) = 0, gives = (3+x-4x^4)' = 0. So 1-16x^3 = 0 . Or x = (1/16)^(1/3) . So f((1/16)^1/3) is an extreme value.

f"(x) = (1-16x^3)' = -48x^2.

f"((1/16)^(1/3) = -48(1/16)^(2/3) is negative.

Therefore f(1/16)^(1/3)) is the maximum at x= (1/16)^(1/3).

Also f(x) is increasing for f'(x) = 1-16x^3 > 0 when x < (1/16)^(1/3).

f(x) is decreasing for 1-16x^3 < 0 when x > (1/16)^(1/3).

The curve is extending downward idefinitely in 4th and 3rd quadrants as x--> +infinity or x--> - infinity.

We'll have to calculate the limit of the function for a specific point of the variable.

We'll calculate the limit for x->0 with x<0. This limit is called the left limit.

lim 3+x-4x^4

We'll substitute x by 0 in the expression of the function:

lim 3+x-4x^4 = 3 + 0 - 4*0^4

lim 3+x-4x^4 = 3

We'll calculate the limit for x->0 with x>0. This limit is called the right limit.

We'll substitute x by 0 in the expression of the function:

lim 3+x-4x^4 = 3 + 0 - 4*0^4

lim 3+x-4x^4 = 3

We notice that the value for the left limit is equal to the value for the right limit.

If we'll calculate the value of the function for x = 0

f(0) = 3

Since the values of both lateral limits are equal to the value of the function for x = 0, the function is continuous for x = 0.