# Question Video: Finding an Unknown in a Rational Function given Its Parity Mathematics • 12th Grade

Find the value of π given π is an even function, where π(π₯) = 6/(8π₯Β² + ππ₯ β 3) and π₯ β  0.

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### Video Transcript

Find the value of π given π is an even function, where π of π₯ equals six over eight π₯ squared plus ππ₯ minus three and π₯ is not equal to zero.

Weβll begin by thinking about what we know about even functions. Letβs imagine we have two functions π sub one and π sub two that are even and π sub two is not zero. Then the quotient π sub one over π sub two must also be even. And secondly, we know that a function is said to be even if π of negative π₯ equals π of π₯ for all values of π₯ in the functionβs domain. Letβs define π sub one to be the numerator of our expression. Thatβs six. Now, in fact, this expression is always equal to six no matter the value of π₯; itβs entirely independent of π₯. And so we can say that π sub one is even.

So, for our function π of π₯ to be even, we need the denominator of our function π of two to also be even. So we define π of two to be equal to eight π₯ squared plus ππ₯ minus three. And we know that if this is even, π sub two of negative π₯ must be equal to π sub two of π₯. Well, π sub two of negative π₯ is the expression eight times negative π₯ squared plus π times negative π₯ minus three. That simplifies to eight π₯ squared minus ππ₯ minus three. And this must be equal to the original function π sub two of π₯.

Weβll now subtract eight π₯ squared from both sides and add three, leaving us with the expression negative ππ₯ equals ππ₯. Now, in fact, weβre also told that π₯ is not equal to zero, so we can divide through by π₯ itself. And that leaves us with the equation negative π equals π. So what values of π make this equation true? The only way for negative π to be equal to π is if π itself is equal to zero.

So, the value of π given that π is an even function is zero.