Solve y=1-x^2/2!+x^4/4!-x^6/6!

$$y = 1 - \frac { x ^ { 2 } } { 2 ! } + \frac { x ^ { 4 } } { 4 ! } - \frac { x ^ { 6 } } { 6 ! }$$

Solve for y

$$y=-\frac{x^{6}}{720}+\frac{x^{4}}{24}-\frac{x^{2}}{2}+1$$

Short Solution Steps

The factorial of 2 is 2.

$$y=1-\frac{x^{2}}{2}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}$$

The factorial of 4 is 24.

$$y=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{6!}$$

The factorial of 6 is 720.

$$y=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\frac{x^{6}}{720}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 24 is 24. Multiply \frac{x^{2}}{2} times \frac{12}{12}.

$$y=1-\frac{12x^{2}}{24}+\frac{x^{4}}{24}-\frac{x^{6}}{720}$$

Since -\frac{12x^{2}}{24} and \frac{x^{4}}{24} have the same denominator, add them by adding their numerators.

$$y=1+\frac{-12x^{2}+x^{4}}{24}-\frac{x^{6}}{720}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 24 and 720 is 720. Multiply \frac{-12x^{2}+x^{4}}{24} times \frac{30}{30}.

$$y=1+\frac{30\left(-12x^{2}+x^{4}\right)}{720}-\frac{x^{6}}{720}$$

Since \frac{30\left(-12x^{2}+x^{4}\right)}{720} and \frac{x^{6}}{720} have the same denominator, subtract them by subtracting their numerators.

$$y=1+\frac{30\left(-12x^{2}+x^{4}\right)-x^{6}}{720}$$

Do the multiplications in 30\left(-12x^{2}+x^{4}\right)-x^{6}.

$$y=1+\frac{-360x^{2}+30x^{4}-x^{6}}{720}$$

Divide each term of -360x^{2}+30x^{4}-x^{6} by 720 to get -\frac{1}{2}x^{2}+\frac{1}{24}x^{4}-\frac{1}{720}x^{6}.

$$y=1-\frac{1}{2}x^{2}+\frac{1}{24}x^{4}-\frac{1}{720}x^{6}$$

Solve for x

$$x=\sqrt{2\left(\sqrt[3]{5\left(3\sqrt{9y^{2}+22y+14}-9y-11\right)}+\sqrt[3]{5\left(-3\sqrt{9y^{2}+22y+14}-9y-11\right)}+5\right)}$$

$$x=-\sqrt{2\left(\sqrt[3]{5\left(3\sqrt{9y^{2}+22y+14}-9y-11\right)}+\sqrt[3]{5\left(-3\sqrt{9y^{2}+22y+14}-9y-11\right)}+5\right)}\text{, }8\sqrt[3]{5}\left(\sqrt[3]{3\sqrt{9y^{2}+22y+14}-9y-11}+\sqrt[3]{-3\sqrt{9y^{2}+22y+14}-9y-11}+5^{\frac{2}{3}}\right)\geq 0$$