Functions #1

Calculus Level pending

Find the domain of the following function:

3 y + 2 x 4 = 2 4 x 2 − 1 \large 3^y+2^{x^4}=2^{4x^2-1}

If your answer is of the form :

( − m − 1 n , − m + 1 n ) ∪ ( m − 1 n , m + 1 n ) (\dfrac{-\sqrt{m}-1}{\sqrt{n}},\dfrac{-\sqrt{m}+1}{\sqrt{n}}) \cup (\dfrac{\sqrt{m}-1}{\sqrt{n}},\dfrac{\sqrt{m}+1}{\sqrt{n}})

Enter : m + n m+n


The answer is 5.

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1 solution

It is easy to see that : 2 4 x 2 − 1 − 2 x 4 > 0 2^{4x^2-1}-2^{x^4}>0

2 4 x 2 − 1 − x 4 > 1 ⟹ 4 x 2 − 1 − x 4 > 0 2^{4x^2-1-x^4}>1 \implies 4x^2-1-x^4>0

x 4 − 4 x 2 + 1 < 0 ⟹ ( x 2 − 2 ) 2 < 3 x^4-4x^2+1 < 0 \implies (x^2-2)^2<3

− 3 + 2 < x 2 < 3 + 2 -\sqrt{3}+2 < x^2 < \sqrt{3}+2

4 − 2 3 2 < x 2 < 4 + 2 3 2 \dfrac{4-2\sqrt{3}}{2}<x^2<\dfrac{4+2\sqrt{3}}{2}

( 2 − 3 ) 2 2 < x 2 < ( 2 + 3 ) 2 2 \dfrac{(2-\sqrt{3})^2}{2}<x^2<\dfrac{(2+\sqrt{3})^2}{2}

Solving the two conditions separately , we get :

x ∈ ( − 3 − 1 2 , − 3 + 1 2 ) ∪ ( 3 − 1 2 , 3 + 1 2 ) x \in (\dfrac{-\sqrt{3}-1}{\sqrt{2}},\dfrac{-\sqrt{3}+1}{\sqrt{2}}) \cup (\dfrac{\sqrt{3}-1}{\sqrt{2}},\dfrac{\sqrt{3}+1}{\sqrt{2}})

Hence , m = 3 \boxed{m=3} and n = 2 \boxed{n=2}

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