Answer to Question #181496 in Calculus for kayci

Question #181496

Prove that 𝑓(π‘₯) = π‘₯^2 + 2π‘₯ is not injective.


1
Expert's answer
2021-05-02T05:13:54-0400

We have , "f(x)=x^2+2x"

"Calculate f(x_1):\\\\\n\n\\Rightarrow f(x_1)=x_1^2+2x\\\\\n\nCalculate f(x_2):\\\\\n\n\\Rightarrow f(x_2)=x_2^2+2x"

Suppose the function is injective then "f(x_1)=f(x_2)"

if this condition is not true then function is not injective

Now, "f(x_1)=f(x_2)"

"\\Rightarrow x_1^2+2x=x_2^2\u200b+2x\u200b"

"\\Rightarrow x_1^2-x_2^2\u200b+2x\u200b_1-2x_2=0\\\\\n\n\\Rightarrow (x_1\u2212x_2)(x_1\u200b+x_2)+2x_1\u22122x_2=0\\\\\n\n\\Rightarrow (x_1\u200b\u2212x_2)(x_1\u200b+x_2+2)=0\\\\"

Since, "x_1\\neq x_2" and "x_1\u200b+x_2\u200b+2\\neq0 ," for any "x\u2208N"

So function f(x) is not injective.


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