3. sketch the graph of y = x^4 - 5x^2 - 14. identify the intervals on which it is increasing, decreasing…

Explanation:

Step1: Find the first - derivative

Differentiate $y = x^{4}-5x^{2}-14$ using the power rule. If $y = x^{n}$, then $y^\prime=nx^{n - 1}$. $y^\prime=4x^{3}-10x=2x(2x^{2}-5)$

Step2: Find critical points

Set $y^\prime = 0$. $2x(2x^{2}-5)=0$ $2x = 0$ gives $x = 0$; $2x^{2}-5=0$ gives $x=\pm\frac{\sqrt{10}}{2}$

Step3: Determine intervals of increase and decrease

Test intervals $(-\infty,-\frac{\sqrt{10}}{2})$, $(-\frac{\sqrt{10}}{2},0)$, $(0,\frac{\sqrt{10}}{2})$, $(\frac{\sqrt{10}}{2},\infty)$ For $x\in(-\infty,-\frac{\sqrt{10}}{2})$, let $x=-2$, $y^\prime=2(-2)[2(-2)^{2}-5]=-4(8 - 5)=-12<0$, so decreasing. For $x\in(-\frac{\sqrt{10}}{2},0)$, let $x = - 1$, $y^\prime=2(-1)[2(-1)^{2}-5]=-2(2 - 5)=6>0$, so increasing. For $x\in(0,\frac{\sqrt{10}}{2})$, let $x = 1$, $y^\prime=2(1)[2(1)^{2}-5]=2(2 - 5)=-6<0$, so decreasing. For $x\in(\frac{\sqrt{10}}{2},\infty)$, let $x = 2$, $y^\prime=2(2)[2(2)^{2}-5]=4(8 - 5)=12>0$, so increasing. Local minimum at $x=\pm\frac{\sqrt{10}}{2}$, $y=(\frac{\sqrt{10}}{2})^{4}-5(\frac{\sqrt{10}}{2})^{2}-14=\frac{100}{16}-\frac{50}{4}-14=\frac{100 - 200 - 224}{16}=-\frac{324}{16}=-\frac{81}{4}$ Local maximum at $x = 0$, $y=-14$

Step4: Find the second - derivative

Differentiate $y^\prime=4x^{3}-10x$ $y^{\prime\prime}=12x^{2}-10 = 2(6x^{2}-5)$

Step5: Find inflection points

Set $y^{\prime\prime}=0$ $2(6x^{2}-5)=0$ $x^{2}=\frac{5}{6}$, so $x=\pm\frac{\sqrt{30}}{6}$

Step6: Determine concavity

Test intervals $(-\infty,-\frac{\sqrt{30}}{6})$, $(-\frac{\sqrt{30}}{6},\frac{\sqrt{30}}{6})$, $(\frac{\sqrt{30}}{6},\infty)$ For $x\in(-\infty,-\frac{\sqrt{30}}{6})$, let $x=-1$, $y^{\prime\prime}=12(-1)^{2}-10 = 2>0$, so concave up. For $x\in(-\frac{\sqrt{30}}{6},\frac{\sqrt{30}}{6})$, let $x = 0$, $y^{\prime\prime}=-10<0$, so concave down. For $x\in(\frac{\sqrt{30}}{6},\infty)$, let $x = 1$, $y^{\prime\prime}=12(1)^{2}-10 = 2>0$, so concave up.

Intervals of increase: $(-\frac{\sqrt{10}}{2},0)\cup(\frac{\sqrt{10}}{2},\infty)$ Intervals of decrease: $(-\infty,-\frac{\sqrt{10}}{2})\cup(0,\frac{\sqrt{10}}{2})$ Intervals of concavity up: $(-\infty,-\frac{\sqrt{30}}{6})\cup(\frac{\sqrt{30}}{6},\infty)$ Intervals of concavity down: $(-\frac{\sqrt{30}}{6},\frac{\sqrt{30}}{6})$ Local minima: $(\pm\frac{\sqrt{10}}{2},-\frac{81}{4})$ Local maximum: $(0,-14)$ Inflection points: $(\pm\frac{\sqrt{30}}{6},(\frac{\sqrt{30}}{6})^{4}-5(\frac{\sqrt{30}}{6})^{2}-14)=(\pm\frac{\sqrt{30}}{6},\frac{900}{1296}-\frac{150}{36}-14)=(\pm\frac{\sqrt{30}}{6},\frac{900 - 5400 - 186624}{1296})=(\pm\frac{\sqrt{30}}{6},-\frac{191124}{1296})$

Answer:

Intervals of increase: $(-\frac{\sqrt{10}}{2},0)\cup(\frac{\sqrt{10}}{2},\infty)$; Intervals of decrease: $(-\infty,-\frac{\sqrt{10}}{2})\cup(0,\frac{\sqrt{10}}{2})$; Intervals of concavity up: $(-\infty,-\frac{\sqrt{30}}{6})\cup(\frac{\sqrt{30}}{6},\infty)$; Intervals of concavity down: $(-\frac{\sqrt{30}}{6},\frac{\sqrt{30}}{6})$; Local minima: $(\pm\frac{\sqrt{10}}{2},-\frac{81}{4})$; Local maximum: $(0,-14)$; Inflection points: $(\pm\frac{\sqrt{30}}{6},-\frac{191124}{1296})$