Differentiable function $f(x)=4x^7 −14x^4 +30x−17$ I am trying to prove that the function $f:\Bbb R→\Bbb R$, $f(x)=4x^7 −14x^4 +30x−17$,is injective. To do this I need to prove it is differentiable from first principles. I can then prove its derivative is strictly increasing to show it is injective. Any help on the proof that it's differentiable would be great, especially in what delta to choose.
 A: Since we can  evaluate a derivative, our function is differentiable. 
Also, $$f'(x)=28x^6-56x^3+30=28\left(x^6-2x^3+\frac{15}{14}\right)=28\left((x^3-1)^2+\frac{1}{14}\right)>0.$$
$$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=$$
$$=\lim_{h\rightarrow0}\frac{4((x+h)^7-x^7)-14((x+h)^4-x^4)+30(x+h-x)}{h}=$$
$$=4(7x^6)-14(4x^3)+30=28x^6-56x^3+30.$$
A: Polynomials are always continuous and differentiable $\forall x \in R$
Here use $\frac{d x^n}{dx}=n x^{n-1}$, Also use the distributivity of $\frac{d}{dx}$ as :$\frac{d}{dx}[ a f(x)+ b g(x)] =a\frac{d}{dx} f(x)+ b\frac{d}{dx} g(x)$, to get
$$\frac{d}{dx}(4x^7-14x^4+30 x-17)= 4 \frac{d}{dx} x^7-14 \frac{d}{dx} x^4 +30\frac{d}{dx}x- \frac{d}{dx}17= 28 x^6-56 x^3 +30. $$
A: Let $f(x)=4x^7-14x^4+30x-17$. Then $$f'(x)=28x^6-42x^3+30=28U^2-42U+30,$$
where $U=x^3$.
Since the discriminant of $28U^2-42U+30$ is negative, the sign of $f'$ won't change. 
A: The OP is asking for a first-principles derivation of the derivative  of a specific polynomial. Here's one way to do it, using the algebraic identity $(x^n-y^n)=(x-y)(x^{n-1}+x^{n-2}y^2+\cdots+xy^{n-2}+y^{n-1})$.
If $f(x)=4x^7-14x^4+30x-17$, then
$$\begin{align}
{f(x)-f(a)\over x-a}
&={4(x^7-a^7)-14(x^4-a^4)+30(x-a)\over x-a}\\
&={4(x-a)(x^6+x^5a+x^4a^2+x^3a^3+x^2a^4+xa^5+a^6)-14(x-a)(x^3+x^2a+xa^2+a^3)+30(x-a)\over x-a}\\
&=4(x^6+x^5a+x^4a^2+x^3a^3+x^2a^4+xa^5+a^6)-14(x^3+x^2a+xa^2+a^3)+30
\end{align}$$
hence
$$\begin{align}
f'(a)&=\lim_{x\to a}{f(x)-f(a)\over x-a}\\
&=\lim_{x\to a}(4(x^6+x^5a+x^4a^2+x^3a^3+x^2a^4+xa^5+a^6)-14(x^3+x^2a+xa^2+a^3)+30)\\
&=4(a^6+a^5a+a^4a^2+a^3a^3+a^2a^4+aa^5+a^6)-14(a^3+a^2a+aa^2+a^3)+30\\
&=4\cdot7a^6-14\cdot4a^3+30\\
&=28a^6-56a^3+30
\end{align}$$
