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Algebra Level 5

Consider a monic polynomial, $f(x)$ , of degree 5. The graphs of $|f(x)|$ and $f(|x|)$ are the same. If 4 is a root of $f(x)$ , then find $f(1)$ .

The answer is 225.

2 solutions

Arjen Vreugdenhil
Apr 27, 2016

The function must be odd, so that $f(0) = 0$ .

If $f(|x|) = |f(x)|$ then $f(x) > 0$ for all positive $x$ .

We are told that $f(4) = 0$ but because $f(x)$ does not become negative in the neighborhood of $x = 4$ , this must be a root of even multiplicity. Because of symmetry, $-4$ is also a root of even multiplicity.

Now we know at least five roots: $-4, -4, 0, 4, 4$ . There cannot be any other. Thus $f(x) = a(x+4)^2\:x\:(x-4)^2.$ Since the polynomial is monic, $a = 1$ . Finally $f(1) = (1+4)^2\cdot 1\cdot (1-4)^2 = 25\cdot 1\cdot 9 = \boxed{225}.$

If a function is odd then how it can be always positive ????

Coz odd function is the reflection w.r.t. X axis

Kushal Bose - 4 years, 11 months ago

It is not positive everywhere. It is only positive for $x > 0$ , and negative for $x < 0$ .

Arjen Vreugdenhil - 4 years, 11 months ago

But there may be another two roots say a & -a.

Kushal Bose - 4 years, 11 months ago

@Kushal Bose – Yes, -4 is also a root with even multiplicity, precisely because the function is odd.

Arjen Vreugdenhil - 4 years, 11 months ago

I now see why my solution is confusing: $f(4) = 0$ is a root of even multiplicity because $f(x)$ does not become negative in the neighborhood of x = 4. My apologies for not stating that clearly.

Arjen Vreugdenhil - 4 years, 11 months ago

Aakash Khandelwal
Apr 22, 2016

If there graph are to be same

$f(x)=(x-4)^{2} (x+4)^{2} x$ .

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The answer is 225.

2 solutions

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