Rolles Theorem, showing two distinct points.

[srhjnmrg]

Have the following question and just wondering if my solution is correct

Let g(x)= x^5+3x-1. Show that there are no distinct points x_1, x_2 in R such that g(x_1)=g(x_2).

Proof by contradiction. Assume we have two solution x_1<x_2 in ℝ, i,e g(x_1)+g(x_2)=0, since g is differentiable on (x_1,x_2) and continuous on [x_1,x_2], then we can apply rolles theorem, there exits a C belonging to (x_1, x_2) such that df/dx=0, however df/dx=5x^4+3>0 Hence we have a contradiction and only one solution to f(x)=0.

Many thanks in advance.

 

[micromass]

Sounds correct!

 

[HallsofIvy]

Have the following question and just wondering if my solution is correct

Let g(x)= x^5+3x-1. Show that there are no distinct points x_1, x_2 in R such that g(x_1)=g(x_2).

Proof by contradiction. Assume we have two solution x_1<x_2 in ℝ, i,e g(x_1)+g(x_2)=0,

I presume you mean g(x_1)- g(x_2)= 0, not the sum.

since g is differentiable on (x_1,x_2) and continuous on [x_1,x_2], then we can apply rolles theorem, there exits a C belonging to (x_1, x_2) such that df/dx=0, however df/dx=5x^4+3>0 Hence we have a contradiction and only one solution to f(x)=0.

Many thanks in advance.

 

[Mark44]

Or you could just take the direct approach. Since g'(x) = 5x⁴ + 3, we see that g'(x) > 0 for all x, which means that the graph of g is increasing on the entire real line, hence g is one-to-one. This fact implies that if g(x₁) = g(x₂), then x₁ = x₂.