Determine the image of a function using **induction**

Question

Question:
Let $f:{\mathbb{R}}\rightarrow{\mathbb{R}}$ by $x \mapsto x^2 + 4x + 7.$ Without using Calculus, show that $Im(f) = [3,\infty)$.

I believe I should prove this by induction, but I'm not sure where to go from there. Thanks.

Attempt 0:

We can see that $I_m(f) = [3,\infty)$ using simple substitution, which yields the domain, $D_f$, as $x = \{\ldots, -3,-2,-1,0,1,2,3,\ldots\}$, and the image, $I_m(f)$, as $y = {}\ldots, 4,3,3,7,12,19,28,\ldots\}$ Thus the image of $f$, $I_m(f) = [3,\infty)$.

Attempt 1: Let $y \in \mathbb{R}$ and let there exists an $x$ such that $f(x) = y$, that is,

$ \begin{eqnarray} x^2 + 4x + 7 &= y \\ \Rightarrow x^2 + 4x + (7-y) &= 0 \\ \Rightarrow \frac{-4 \pm \sqrt{4^2 - 4 (7-y)}}{2} &= x \\ \Rightarrow 4^2 - 4(7-y) &\geq 0 \\ \Rightarrow y &\geq 3. \end{eqnarray}$

Conversely, when $y < 3$ there is no solution. Therefore, $Im(f) = [3,\infty)$.

Answer 1

Just write $$x^2+4x+7=(x+2)^2+3\geq3.$$

Answer 2

Your answer is wrong. The set $D_f$ is $\mathbb R$; you cannot change that. It happens that $f'(x)=2x+4$. Therefore, $f$ decreases on $(-\infty,-2]$ and increases on $[-2,+\infty)$. So, $\min f=3$ and $\operatorname{Im}f\subset[3,+\infty)$. Since $\lim_{x\to\pm\infty}f(x)=+\infty$ it follows from the intermediate value theorem that $\operatorname{Im}f$ is $[3,+\infty)$.

Answer 3

$$x^2+4x+7=y \iff x^2+4x+(7-y)=0 \iff x = \dfrac {-4 \pm \sqrt {4^2-4(7-y)}}{2} $$

$$\iff 4^2-4(7-y) \ge 0 \iff y \ge 3$$

This shows that for $y \in \mathbb{R}$, there exists an $x$ such that $f(x)=y \iff y \ge 3$, which tells us that the range of $f$ is $[3, \infty)$

Answer 4

Let $u \in \mathbb R$.

1. Show that the quadratic equation $x^2+4x+7=u$ has a solution , if $u \ge 3$.

2. Show that the quadratic equation $x^2+4x+7=u$ has no solution , if $u <3 $.