Show that $h(x)=x^5+3x+6$ is one to one

Question

How do I show that $$h(x)=x^5+3x+6$$ is one to one? I set $$f(a)=f(b)$$ and try to isolate for $a$ and $b$ but I get stuck because I have a term of "$a$" that is degree $5$ and a term of "$a$" that is degree $1$ so I am unable to simplify it.

Answer 1

Hint: $h'(x) = 5x^4 + 3 > 0$

Answer 2

Obviously $h$ is increasing iff $h-6$ is increasing. Let be $$g(x)=h(x)-6=x(x^4+3).$$ As $g$ is odd, is enough to prove that $g$ is increasing in $\Bbb R^+$.

Is easy to see that the product of increasing positive functions is increasing: $$x<y\implies f_1(x)f_2(x)<f_1(x)f_2(y)<f_1(y)f_2(y).$$ Finally, $$x\mapsto x\text{ is positive and increasing in $\Bbb R^+$},$$ $$x\mapsto x^4\text{ is positive and increasing in $\Bbb R^+$},$$ $$x\mapsto x^4+3\text{ is positive and increasing in $\Bbb R^+$},$$ $$x\mapsto x(x^4+3)\text{ is positive and increasing in $\Bbb R^+$}.$$

Answer 3

Since you asked (in a comment on Integrator's answer) for a proof without using the derivative: We have to show that $a\lt b\implies h(a)\lt h(b)$. Since $a\lt b\implies 3a+6\lt3b+6$, it will suffice to show that $a\lt b\implies a^5\lt b^5$. We can do it by cases.

I. If $0\le a\lt b$ then $a^5\le a^4b\le a^3b^2\le a^2b^3\le ab^4\lt b^5$; alternatively, $b^5-a^5=(b-a)(b^4+ab^3+a^2b^2+a^3b+a^4)\gt0$.

II. If $a\lt0\lt b$ then $a^5\lt0\lt b^5$.

III. If $a\lt b\lt0$ then $0\lt-b\lt-a$, so $-b^5=(-b)^5\lt(-a)^5=-a^5$ by case I, and so $a^5\lt b^5$.

Answer 4

Hint: $$f(a)=f(b)$$ $$f(a)-f(b)=0$$

$$a^5-b^5+3a-3b=0$$ $$(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4+3)=0$$

Answer 5

NOT a hint:

It may be shown that

$h(x) = x^5 + 3x + 6 \tag{1}$

is one-to-one without introducing derivatives by noting that it is a strictly monotonically increasing function. That is, if $x_1 > x_2$, then

$h(x_1) = x_1^5 + 3x_1 + 6 > x_2^5 + 3x_2 + 6 = h(x_2). \tag{2}$

(2) holds since both $x^5$ and $3x$ are themselves strictly monotonically increasing: for $x_1 > x_2$, we have

$x_1^5 > x_2^5 \tag{3}$

and

$3x_1 > 3x_2. \tag{4}$

Adding (3) and (4) yields

$x_1^5 + 3x_1 > x_2^5 + 3x_2, \tag{5}$

whence follows (2) post haste. It is now evident that $h(x)$ is one-to-one, since if $x_1 \ne x_2$, either $x_1 > x_2$ or vice versa, etc. etc. etc. There it is, no calculus!!!

Hope this helps. Cheers,

and as ever,

Fiat Lux!!!

Answer 6

$$f'(x)=5x^4+3$$ which is positive. Therefore f(x) is monotonically increasing.

Therefore: It is one-on-one . Since if $x\ne y$ It is either $x<y$ or $x>y$ then either $f(x)< f(y)$ or $f(x)>f(y)$ by monotonicity and hence $f(x)\ne f(y)$.