We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
276
2
avatar+75 

If f(x)=x10+2x9-2x8-2x7+x6+3x2+6x+1, then what is f(√(2-1))

 Oct 16, 2018
 #1
avatar+8249 
+1

\(If\ f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1\ ,\) 

                                                                        \(then\ what\ is\ f(\sqrt{2-1}\ )\)

 

\(f(\sqrt{2-1})=f(\sqrt1\ )=1\)

 

laugh  !

 Oct 16, 2018
edited by asinus  Oct 16, 2018
 #2
avatar+322 
+2

Why f(1) = 1? 

\(f(\sqrt{2-1}) = f(\sqrt{1}) = f(1) \)

so \(f(1)= 1^10 + 2(1)^9 - 2(1)^8 - 2(1)^7 + 1^6 + 3(1)^2 + 6(1) + 1 = 1 + 2 - 2 - 2 + 1 + 3 + 6 + 1 = 3 + 6 + 1 = 10 \)

so \(f(1) = 10 \) not 1,where im wrong?

Thanks! 

Dimitristhym  Oct 16, 2018
edited by Dimitristhym  Oct 16, 2018

13 Online Users

avatar