Answer to Question #181496 in Calculus for kayci

Question #181496

Prove that 𝑓(𝑥) = 𝑥^2 + 2𝑥 is not injective.


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

We have , f(x)=x2+2xf(x)=x^2+2x

Calculatef(x1):f(x1)=x12+2xCalculatef(x2):f(x2)=x22+2xCalculate f(x_1):\\ \Rightarrow f(x_1)=x_1^2+2x\\ Calculate f(x_2):\\ \Rightarrow f(x_2)=x_2^2+2x

Suppose the function is injective then f(x1)=f(x2)f(x_1)=f(x_2)

if this condition is not true then function is not injective

Now, f(x1)=f(x2)f(x_1)=f(x_2)

x12+2x=x22+2x\Rightarrow x_1^2+2x=x_2^2​+2x​

x12x22+2x12x2=0(x1x2)(x1+x2)+2x12x2=0(x1x2)(x1+x2+2)=0\Rightarrow x_1^2-x_2^2​+2x​_1-2x_2=0\\ \Rightarrow (x_1−x_2)(x_1​+x_2)+2x_1−2x_2=0\\ \Rightarrow (x_1​−x_2)(x_1​+x_2+2)=0\\

Since, x1x2x_1\neq x_2 and x1+x2+20,x_1​+x_2​+2\neq0 , for any xNx∈N

So function f(x) is not injective.


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