Show that $f(x)=2x^2+4x+5$ is positive for all real values of $x$.

Question

• Show that the function $f(x)=2x^2+4x+5$ is positive for all real values of $x$.

  At first I used completing the square technique, ${ax^2 + bx + c}$ is converted to ${a(x + h)^2 + k}$

  $2x^2+4x+5$

  = ${2(x^2+2x+5/2)}$

  = ${2((x+1)^2+3/2)}$

  = ${2(x+1)^2+3}$

  since ${(x+1)^2}$ is always positive and the coefficient of ${x^2}$ is 2>0, thus ${f(x)>0}$

• And find its minimum value.

  Again I used completing the square technique to find it's minimum value : 3

  ${2(x+1)^2+3}$ ≥ ${3}$

• Hence show that $0<\frac{6}{f(x)}\leq2$.

  but at last I've got trouble with this, If anyone can help, I appreciate it

Answer 1

Hints:

1. Try to complete the square.

2. Set the derivative equal to zero or use the formula that the minimum of $ax^2+bx+c$ is attained at $x=-\frac{b}{2a}$.

3. Use (2) and because $f(x)$ is positive, you may divide by $f(x)$.

Answer 2

1. $\Delta=16-40<0$ and the coefficient of $x^2$ is $2>0$, thus $f(x)>0\;\forall x\in\Bbb R$.
2. $f'(x)=4x+4$ is equal to $0$ iff $x=-1$ thus $\min f=f(-1)=2-4+5=3$
3. from previous point $f(x)\ge3\;\forall x\in\Bbb R$, thus $0\le\frac1{f(x)}\le\frac13$ and multiply all three sides by $6$ your result follows.

Answer 3

1) $f(x)=2x^2+4x+5=2x^2+4x+2+3=2(x^2+2x+1)+3=2(x+1)^2+3>0$

2) $f(x)=2(x+1)^2+3 \ge 3$ and if $x=-1$ that $f(-1)=3$

3) If $f(x) \ge 3$ that $$0<\frac{1}{f(x)}\le \frac 13$$ $$0<\frac{6}{f(x)}\le 2$$