Prelim 2016 Maths Help Thread (1 Viewer)

davidgoes4wce · Jun 12, 2016

eyeseeyou said:
what is the domin of -x+ square root of (x+3)?

I got x is greater than or equal to -3 (Domain) and y is greater than or equal to 0 (range)

$f(x)=-x+\sqrt{x+3}$

$f'(x)=-1+\frac{1}{2} (x+3)^{\frac{-1}{2}}$

$f'(x)=0$

$0=-1+\frac{1}{2} (x+3)^{\frac{-1}{2}}$

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

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

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

$Take the square of both sides$

$(x+3)=\frac{1}{4}$

$x=\frac{1}{4}-3=\frac{-11}{4}$

$$ The turning point occurs at $ x=\frac{-11}{4}. $ Substituting x $ =\frac{-11}{4} $ into f(x), yields $ f(\frac{-11}{4})=-(-\frac{11}{4})+\sqrt{\frac{-11}{4}+3}=\frac{11}{4}+\sqrt{\frac{1}{4}}=\frac{11}{4}+\frac{1}{2}=\frac{13}{4}=3.25$

$The value of 3.25 or $ \frac{13}{4} $ happens to be the highest or maximum value for the function f(x) [Graph Below]$

$$ The Domain of x is where the square root $ \sqrt{x+3} $ the term inside the bracket has to be $ \geq 0$

$x+3 \geq 0$

$x \geq -3$

$$ Domain of x, $ \{ x \in \mathbb{R} : x \geq -3 \}$

$$ Range of f(x), $\{y \in \mathbb{R} : y \leq \frac{13}{4} \}$

eyeseeyou · Jun 12, 2016

davidgoes4wce said:
$f(x)=-x+\sqrt{x+3}$

$f'(x)=-1+\frac{1}{2} (x+3)^{\frac{-1}{2}}$

$f'(x)=0$

$0=-1+\frac{1}{2} (x+3)^{\frac{-1}{2}}$

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

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

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

$Take the square of both sides$

$(x+3)=\frac{1}{4}$

$x=\frac{1}{4}-3=\frac{-11}{4}$

$$ The turning point occurs at $ x=\frac{-11}{4}. $ Substituting x $ =\frac{-11}{4} $ into f(x), yields $ f(\frac{-11}{4})=-(-\frac{11}{4})+\sqrt{\frac{-11}{4}+3}=\frac{11}{4}+\sqrt{\frac{1}{4}}=\frac{11}{4}+\frac{1}{2}=\frac{13}{4}=3.25$

$The value of 3.25 or $ \frac{13}{4} $ happens to be the highest or maximum value for the function f(x) [Graph Below]$

$$ The Domain of x is where the square root $ \sqrt{x+3} $ the term inside the bracket has to be $ \geq 0$

$x+3 \geq 0$

$x \geq -3$

$$ Domain of x, $ \{ x \in \mathbb{R} : x \geq -3 \}$

$$ Range of f(x), $\{y \in \mathbb{R} : y \leq \frac{13}{4} \}$

Do not include any calculus in this

Trebla · Jun 12, 2016

eyeseeyou said:
Yes

Do you think I'm now ready for 4U?

So what was the problem you had with the question originally?

integral95 · Jun 12, 2016

eyeseeyou said:
Yes

Do you think I'm now ready for 4U?

alright this confirms it, he's a complete troll.

parad0xica · Jun 12, 2016

eyeseeyou said:
Do not include any calculus in this

Why?!

Paradoxica · Jun 12, 2016

eyeseeyou said:
Do not include any calculus in this

By inspection the maximum value of the expression is 13/4.

Hey, you said no calculus, inspection was never excluded.

parad0xica · Jun 12, 2016

Paradoxica said:
By inspection the maximum value of the expression is 13/4.

Hey, you said no calculus, inspection was never excluded.

http://imgur.com/QoGpzut

Paradoxica · Jun 12, 2016

Now, for a more serious solution.....

make the substitution x = y-3

The function, in terms of y

3 - y + √y

Complete the square:

13⁄4 - (1⁄2-√y)²

back substitute to return x

13⁄4 - (1⁄2-√(x+3))²

The maximum value of the expression occurs when the negative term is as small as possible (in this case, 0)

So the maximum value is 13⁄4

The minimum value occurs when the negative term is as large as possible. Since there is no upper bound on this term, the minimum does not exist.

So the function range goes from -∞ to 13⁄4

drsabz101 · Jun 12, 2016

Can someone draw the situation, I know how to draw the two given lines, but not the 'M' and 'n'?

InteGrand · Jun 12, 2016

drsabz101 said:
Can someone draw the situation, I know how to draw the two given lines, but not the 'M' and 'n'? View attachment 33266

$\noindent Draw from $P$ line segments to the given lines that are perpendicular to those lines. The points where they meet the given lines are $M$ and $N$. (The answer to the locus will be the lines bisecting the angles made by the lines.)$

drsabz101 · Jun 12, 2016

I still can't visualise this

InteGrand · Jun 12, 2016

drsabz101 said:
I still can't visualise this

Try just drawing a perpendicular line segment from a random point P to one of the lines, and then to the other line (image: https://www.mathsisfun.com/images/perpendicular.gif ).

drsabz101 · Jun 12, 2016

okay i got it thanks!

davidgoes4wce · Jun 12, 2016

InteGrand said:
$\noindent Draw from $P$ line segments to the given lines that are perpendicular to those lines. The points where they meet the given lines are $M$ and $N$. (The answer to the locus will be the lines bisecting the angles made by the lines.)$

I'm assuming in this question the two lines that go through this , go through the x-coordinate

$x=\frac{-1}{12}$

Below is my graph below:

drsabz101 · Jun 12, 2016

how do u find the locus of p?

InteGrand · Jun 12, 2016

drsabz101 said:
how do u find the locus of p?

It's the pair of intersecting lines that bisect the angles made by the two given lines (they go through the point of intersection of those two lines).

If you want the equations of the lines, the way they'd expect you to do it is to use the "perpendicular distance formula" and then equate the distances of P to each of the two given lines.

drsabz101 · Jun 12, 2016

why do we need to consider a positive and negative case?

InteGrand · Jun 12, 2016

drsabz101 said:
why do we need to consider a positive and negative case?

The locus will be two lines. Both cases need to be considered to get the full locus.

drsabz101 · Jun 12, 2016

can someone draw a diagram labeling the locus, I am not getting this ( i am self-teaching)

InteGrand · Jun 12, 2016

drsabz101 said:
can someone draw a diagram labeling the locus, I am not getting this ( i am self-teaching)

Draw the two lines going through the point of intersection of the given lines and bisecting the angles made by these two given lines. This is the locus.

Prelim 2016 Maths Help Thread (1 Viewer)

Well-Known Member

Well-Known Member

Administrator

Well-Known Member

Active Member

-insert title here-

Active Member

-insert title here-

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)