1. [SOLVED] Function Question....

Question:
Express x^2 + 4x in the form (x + a)^2 + b , stating the numerical values of a and b.

The functions f and g are defined as follows:

$f : x \mapsto x^2 + 4x$,██ $x \geq -2$,

$g : x \mapsto x + 6$,██ $x \in R$

(i) Show that the equation $gf(x) = 0$ has no real roots.
(ii) State the domain of $f^{-1}$, and find an expression in terms of $x$ for $f^{-1}(x)$.
(iii) Sketch, in a single diagram, the graph of $y = f(x)$ and $y = f^{-1}(x)$, making clear the relationship between these graphs.

Attempt:

$x^2 + 4x$

$\Rightarrow \left(x + \frac{4}{2}\right)^2 - \left(\frac{4}{2}\right)^2$

$\Rightarrow (x + 2)^2 - 4$

$a = 2$ , $b = -4$

$(i) gf(x) = (x^2 + 4x) + 6$
$gf(x) = (x + 2)^2 - 4 + 6$
$gf(x) = (x + 2)^2 + 2$

$(x + 2)(x + 2)$
$\Rightarrow x^2 + 2x + 2x + 4$
$\Rightarrow x^2 + 4x + 4$

$gf(x) = x^2 + 4x + 6$
$a = 1, b = 4, c = 6$

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x = \frac{4 \pm \sqrt{4^2 - 4 \times 1 \times 6}}{2 \times 1}$

I can't complete because I get a negative value in the square root. Where did I go wrong?

2. Originally Posted by looi76
Question:
Express x^2 + 4x in the form (x + a)^2 + b , stating the numerical values of a and b.

The functions f and g are defined as follows:

$f : x \mapsto x^2 + 4x$,██ $x \geq -2$,

$g : x \mapsto x + 6$,██ $x \in R$

(i) Show that the equation $gf(x) = 0$ has no real roots.
(ii) State the domain of $f^{-1}$, and find an expression in terms of $x$ for $f^{-1}(x)$.
(iii) Sketch, in a single diagram, the graph of $y = f(x)$ and $y = f^{-1}(x)$, making clear the relationship between these graphs.

Attempt:

$x^2 + 4x$

$\Rightarrow \left(x + \frac{4}{2}\right)^2 - \left(\frac{4}{2}\right)^2$

$\Rightarrow (x + 2)^2 - 4$

$a = 2$ , $b = -4$

$(i) gf(x) = (x^2 + 4x) + 6$
$gf(x) = (x + 2)^2 - 4 + 6$
$gf(x) = (x + 2)^2 + 2$

$(x + 2)(x + 2)$
$\Rightarrow x^2 + 2x + 2x + 4$
$\Rightarrow x^2 + 4x + 4$

$gf(x) = x^2 + 4x + 6$
$a = 1, b = 4, c = 6$

$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x = \frac{4 \pm \sqrt{4^2 - 4 \times 1 \times 6}}{2 \times 1}$

I can't complete because I get a negative value in the square root. Where did I go wrong?
The fact that you get a negative value in the square root illustrates that the function $x^2 + 4x + 6$ has no real roots, which is what you want to show.

3. Thanks icemanfan

(ii) $(y + 2)^2 - 4 = x$
$(y + 2)^2 = x + 4$
$y + 2 = \sqrt{x + 4}$
$y = \sqrt{x + 4} - 2$

$f^{-1}(x) = \sqrt{x + 4} - 2$

What is the domain and what is the range?

4. i hope you know the rules on real and imaginary roots:
ax^2 + b.x + c

if b^2 - 4.a.c > 0 you will have two different real roots
if b^2 - 4.a.c = 0 you will have equal real roots
if b^2 - 4.a.c < 0 you will have imaginary conjugate roots

.
.
.
= (x+2)^2 +2
= x^2 + 4x +6 ====> a=1, b=4, c=6

b^2 - 4.a.c = -8 (two imaginary roots)

5. Originally Posted by looi76
Thanks icemanfan

(ii) $(y + 2)^2 - 4 = x$
$(y + 2)^2 = x + 4$
$y + 2 = \sqrt{x + 4}$
$y = \sqrt{x + 4} - 2$

$f^{-1}(x) = \sqrt{x + 4} - 2$

What is the domain and what is the range?
This is a tricky question. It turns out that you can also find the inverse function $y = -\sqrt{x + 4} - 2$. But for the function
$y = \sqrt{x + 4} - 2$, the domain is $x \geq -4$, all of the numbers that will maintain a nonnegative value within the square root. The range for this function is $y \geq -2$, since the square root is always greater than or equal to zero. See if you can find the domain and range for the other inverse function.

i hope you know the rules on real and imaginary roots:
ax^2 + b.x + c

if b^2 - 4.a.c > 0 you will have two different real roots
if b^2 - 4.a.c = 0 you will have equal real roots
if b^2 - 4.a.c < 0 you will have imaginary conjugate roots

.
.
.
= (x+2)^2 +2
= x^2 + 4x +6 ====> a=1, b=4, c=6

b^2 - 4.a.c = -8 (two imaginary roots)

What does Discriminant mean? In the textbook it is written that "Discriminant is -8 < 0"

7. Originally Posted by looi76

(iii) Sketch, in a single diagram, the graph of $y = f(x)$ and $y = f^{-1}(x)$, making clear the relationship between these graphs.

8. b^2 - 4a.c is called the discriminant.

9. Originally Posted by looi76

What does Discriminant mean? In the textbook it is written that "Discriminant is -8 < 0"
Hello,

Discriminant - Wikipedia, the free encyclopedia

To sum up :

If you have an equation $ax^2+bx+c=0$

$\Delta=b^2-4ac$

If $\Delta <0$, the equation has no real root and $ax^2+bx+c$ is of the same sign as a.

If $\Delta=0$, $ax^2+bx+c=\left(x+\frac{b}{2a}\right)^2$

If $\Delta>0$, $ax^2+bx+c=\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)$

b^2 - 4a.c is called the discriminant.
Originally Posted by Moo
Hello,

Discriminant - Wikipedia, the free encyclopedia

To sum up :

If you have an equation $ax^2+bx+c=0$

$\Delta=b^2-4ac$

If $\Delta <0$, the equation has no real root and $ax^2+bx+c$ is of the same sign as a.

If $\Delta=0$, $ax^2+bx+c=\left(x+\frac{b}{2a}\right)^2$

If $\Delta>0$, $ax^2+bx+c=\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)$