Please help figuring out this function.f(x) = x^4 + 2x^3 - 4x^2 + x - 1

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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The domain of the function is the set of real numbers.

You need to check if there are any horizontal asymptotes of the graph, to `+-oo` .

`lim_(x-gtoo)f(x) = oo^4 + 2oo^3 - 4oo^2 +oo - 1 = oo`

`lim_(x-gt-oo)f(x) = (-oo)^4 + 2(-oo)^3 - 4(-oo)^2-oo - 1 = oo`

There are no horizontal asymptotes of the graph of f(x).

Since the function f(x) exists for all values of x, there are no vertical asymptotes of the graph of f(x).

You need to check if there are any slant asymptotes of the graph, to`+-oo` .

The equation of the slant asymptote is y = mx + n.

`m = lim_(x-gtoo) f(x)/x =gt lim_(x-gtoo) (x^4 + 2x^3 - 4x^2 + x - 1)/x =gt m -gt =oo`

Since m has no finite value, there are no slant asymptotes.

You need to calculate the derivative of the function to check if there are stationary points.

`f'(x) = 4x^3 + 6x^2 - 8x + 1`

Since the derivative has no roots, the function has no stationary points.

Calculate `f"(x) =gt f"(x) = 12x^2 + 12x - 8`

`12x^2 + 12x - 8 = 0 =gt 3x^2 + 3x - 2 = 0`

`` `x_(1,2) = (-3+-sqrt(9 + 24))/6`

`x_(1,2) = (-3+-sqrt33)/6`

The function has inflection points at `x_(1,2) = (-3+-sqrt33)/6` .

The function has no horizontal, vertical or slant asymptotes. The function has no stationary points but it has inflection points at`x_(1,2) = (-3+-sqrt33)/6.`

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