Question

Sketch the graph of the function. Label the intercepts, relative extrema, points of inflection, and asymptotes. Then state the domain of the function. $y=\frac{x^4}{x^4-1}$

Short answer

The function $y=\frac{x^4}{x^4-1}$ has an intercept at the origin (0,0), vertical asymptotes at $x=\pm 1$, and a horizontal asymptote at $y=1$. There are no local extrema or points of inflection. The domain is $x\in R,x\ne \pm 1$.

Step-by-step solution

Identify the Intercept

The x-intercept is found by setting $y=0$, leading to the conclusion that $x^4=0$, therefore $x=0$. The y-intercept is found by setting $x=0$ in the function, giving $y=0$. The only intercept is at (0,0).

Find the Asymptotes

To find vertical asymptotes, set the denominator equal to zero and solve for $x$: $x^4-1=0$ yielding $x^2=\pm 1$, thus $x=\pm 1$ are the vertical asymptotes. The horizontal asymptote is found by taking the limit of the function as $x$ approaches ±∞, which results in $y=1$.

Identify Local Extrema and Inflection Points

Taking the first derivative, $y^{\prime }=\frac{4x^3(x^4-1)-x^4(4x^3)}{(x^4-1{)}^2}=0$, it can be seen that finding the critical points is a complex task and due to the nature of the function no real critical points exist. Therefore, there are neither relative extrema nor points of inflection.

Identify the Domain

Examine the function to identify the domain. The denominator must not equal zero, hence $x^4-1\ne 0$. This leads to the conclusion that $x\ne \pm 1$. Therefore the domain is $x\in R,x\ne \pm 1$