# Thread: [SOLVED] Function Question....

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.

6. Originally Posted by Danshader
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.

Can someone please help me with the diagram? I can't draw it!

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

10. Originally Posted by Danshader
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 $\Delta>0$, $ax^2+bx+c=\left(x-\frac{-b+\sqrt{\Delta}}{2a}\right)\left(x-\frac{-b-\sqrt{\Delta}}{2a}\right)$