# Chapter 4 - Section 4.5 - Summary of Curve Sketching - 4.5 Exercises - Page 321: 3

A. The domain is $(-\infty, \infty)$ B. The y-intercept is $0$ The x-intercepts are $0$ and $\sqrt 4$ C. The function is not odd or even. D. $\lim\limits_{x \to -\infty} (x^4-4x) = \infty$ $\lim\limits_{x \to \infty} (x^4-4x) = \infty$ There are no asymptotes. E. The function is increasing on the intervals $(1, \infty)$ The function is decreasing on the interval $(-\infty,1)$ F. $(1,-3)$ is a local minimum. G. The graph is concave up on the intervals $(-\infty, 0) \cup (0, \infty)$ There is no point of inflection. H. We can see a sketch of the curve below. $y = x^4-4x$ A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$ B. When $x = 0,$ then $y = (0)^4-4(0) = 0$ The y-intercept is $0$ When $y = 0$: $x^4+4x = 0$ $x(x^3-4) = 0$ $x = 0$ or $x = \sqrt 4$ The x-intercepts are $0$ and $\sqrt 4$ C. The function is not odd or even. D. $\lim\limits_{x \to -\infty} (x^4-4x) = \infty$ $\lim\limits_{x \to \infty} (x^4-4x) = \lim\limits_{x \to \infty} (x)(x^3-4)=\infty$ There are no asymptotes. E. We can find the values of $x$ such that $y' = 0$: $y' = 4x^3-4 = 0$ $x^3-1 = 0$ $x = 1$ The function is increasing on the intervals $(1, \infty)$ The function is decreasing on the interval $(-\infty,1)$ F. When $x = 1$, then $y = (1)^4-4(1)= -3$ $(1,-3)$ is a local minimum. G. We can find the values of $x$ such that $y'' = 0$: $y'' = 12x^2 = 0$ $x = 0$ When $x \lt 0$ or $x \gt 0$, then $y'' \gt 0$ The graph is concave up on the intervals $(-\infty, 0) \cup (0, \infty)$ There is no point of inflection since the concavity does not change at $x=0$ H. We can see a sketch of the curve below. 