If f(x)=x^2 g(x)=x^4?

Question

f * g(x)= x^2 * x^4=x^6
how do u prove that itz even

Answer 1

they're only even if x is 0, -1, or 1

whats f * g (x) supposed to be in the second line?

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oh nevermind this... i thought u meant 'equal' not even.

Answer 2

If you mean f(x)*g(x) = x^2*x^6=x^6, and by "even" you mean both sides of the equation are equal? Then, just substitute some random number into x, and solve. For example, use 2 for x. So, 2^2*2^4 should equal to 2^6, which they are.

Answer 3

Show the case using an arbitrary variable where if x is odd then f*g has to be even, and do the same thing using an even number

something like

x = 2*a even
x = 2*a+1 odd

Answer 4

Simple-- substitute -x for x in the equation. If it comes out the same, then the function is even. In this case, they might want you to start from x^2-- if you can prove that that's even, then you can prove that x^4 is even, because the product of two even functions is even. Then you can use the same rule for x^6...

For example if h(x)=x, then h(-x)=-x, so h is an odd function.
If h(x)=0. then h(-x)=0, so h(x) is even...

Enjoy

Answer 5

I think, you didn't mean to ask to prove the number x^6 to be even, rather you meant to prove that the new function f(x)*g(x) to be even. If that is the case, I can prove that -

Given, f(x) = x^2, g(x) = x^4
So, the new function, say, p(x) = f(x) * g(x) = x^6.

Now, the funtion p(x) is even if p(-x) = p(x) and the function is odd if p(-x) = - p(x).

Here,
p(x) = x^6
so, p(-x) = (-x)^6; [replacing x by -x]
p(-x) = (-x)^6
p(-x) = x^6; [because, (-x)^6 = x^6]
so, p(-x) = p(x).

Thus, p(x) is an even function.

Answer 6

kyo paka raha hai
hame nahi pata! bhaag yahan se!

Answer 7

They aren't!

Answer 8

no!