y=tan^2(1+2x^2)

Lösung

y = tan^{2} (1 + 2 x^{2})

Parity : Gerade

Inverse : \frac{arctan ( x ) - 1}{2}, - \frac{arctan ( x ) - 1}{2}, \frac{arctan ( - x ) - 1}{2}, - \frac{arctan ( - x ) - 1}{2}

Schritte zur Lösung

Parity of

tan^{2} (1 + 2 x^{2}) :

Gerade

Umkehrung von

tan^{2} (1 + 2 x^{2}) : \frac{arctan ( x ) - 1}{2}, - \frac{arctan ( x ) - 1}{2}, \frac{arctan ( - x ) - 1}{2}, - \frac{arctan ( - x ) - 1}{2}

f (x) = 3 \frac{4}{9 x ^{2}}

y = 4 sin (π x)

f (x) = 3 6 x - 5

f (x) = \frac{2 x + 4}{x - 2}

f (x) = ∣ x + 9 ∣