Mathematics Question Thread

[RuiAce]

Hey everyone,

Can somebody show me their working out to the attached question? I want to compare it to mine. The answer claims that the maximum is -2.

It should be. The only stationary point is at \(x=0\), so we only need to check the value of \(y\) when \(x=0\), and those at the endpoints \(-2\) and \(1\). The candidate values are therefore \(-3\), \(-35\) and \(-2\), so the global maximum value is \(-2\).

Note that global maxima/minima do not necessarily coincide with local maxima/minima. They can (and sometimes do in fact) occur on the endpoints of the domain.

[alexnero7]

Thank you!! 

It should be. The only stationary point is at \(x=0\), so we only need to check the value of \(y\) when \(x=0\), and those at the endpoints \(-2\) and \(1\). The candidate values are therefore \(-3\), \(-35\) and \(-2\), so the global maximum value is \(-2\).

Note that global maxima/minima do not necessarily coincide with local maxima/minima. They can (and sometimes do in fact) occur on the endpoints of the domain.

[meerae]

Hey everyone,

Can somebody show me their working out to the attached question? I want to compare it to mine. The answer claims that the maximum is -2.

Hey!

I've attached my working out - if you need any clarification please let me know!

Hope this helped!
meerae

[fun_jirachi]

Hi there!

The simplest way is probably to take a geometrical approach. By rearranging the second equation to y=x+2, and equating (or recognising) that they intersect at (0, 2) and (-2, 0). The area enclosed by the curve is a quarter of a circle (top right quadrant of the circle x^2+y^2=4), plus a triangle enclosed by the coordinate axes and the line y=x+2, that cuts at (0, 2) and (-2, 0). This triangle has base 2 and height 2 (thinking geometrically) and likewise the semicircle given gives you the dimensions of the quadrant that you need to calculate the area for. I got pi+2 using this method.

Hope this helps

[eeshab7]

Hi guys, I've had a complete mind blank and can't remember how to do this question. Could someone please help me.
Thanks

[meerae]

Hi guys, I've had a complete mind blank and can't remember how to do this question. Could someone please help me.
Thanks

Hey!
You would use the product rule to differentiate this. just remember for exponentials that if y= e^2x then y'= 2e^2x
You would then find the x and y intercepts of the graph and possibly inflection points (achieved by differentiating again and making y''=0) and then utilise this information to sketch the graph. Does that make sense?

Hope this helped!
meerae

[eeshab7]

Hey!
You would use the product rule to differentiate this. just remember for exponentials that if y= e^2x then y'= 2e^2x
You would then find the x and y intercepts of the graph and possibly inflection points (achieved by differentiating again and making y''=0) and then utilise this information to sketch the graph. Does that make sense?

Hope this helped!
meerae

Hi,
Thanks for that and yeah it makes sense. This might be a bit messy but am I on the right track?

[RuiAce]

Hi,
Thanks for that and yeah it makes sense. This might be a bit messy but am I on the right track?

\[ \text{Your product rule computation should be}\\ \begin{align*}y&=x^2e^{2x}\\ \frac{dy}{dx} &= (2x)(e^{2x}) + (x^2)(2e^{2x})\\ &= 2e^{2x} x(x+1) \end{align*} \]
Also, are you sure you meant to have two dashes there on your \(f^{\prime\prime}(x)\)?

[eeshab7]

\[ \text{Your product rule computation should be}\\ \begin{align*}y&=x^2e^{2x}\\ \frac{dy}{dx} &= (2x)(e^{2x}) + (x^2)(2e^{2x})\\ &= 2e^{2x} x(x+1) \end{align*} \]
Also, are you sure you meant to have two dashes there on your \(f^{\prime\prime}(x)\)?

yeah sorry I forgot to include the previous steps in the picture - that was just using the double derivative

yeah sorry I forgot to include the previous steps in the picture - that was just using the double derivative

is that right for the next step?

[RuiAce]

is that right for the next step?

Still looks off for the second derivative. Just off by a +1 term.
\begin{align*} \frac{dy}{dx} &= (2x^2+2x)e^{2x}\\ \frac{d^2y}{dx^2} &= (4x+2)e^{2x} + (2x^2+2x)(2e^{2x}) \\ &= (4x+2+4x^2+4x)e^{2x}\\ &= 2(2x^2+4x+1)e^{2x} \end{align*}
Although, if we just want to test the nature of the stationary points, leaving it as \( \frac{d^2y}{dx^2} = (4x+2)e^{2x} + 2(2x^2+2x)e^{2x} \) is fine - no need to simplify.

[alexnero7]

I cant seem to get the right answer for this... it's SO frustrating!! Please HELP!!

[RuiAce]

I cant seem to get the right answer for this... it's SO frustrating!! Please HELP!!

You have \( \frac{d^2y}{dx^2} = 12x - 14\) so all you're trying to solve is \(12x - 14 > 0\), which becomes \(x > \frac76\).

[alexnero7]

Thank you Rui!! I understood it now.

You have \( \frac{d^2y}{dx^2} = 12x - 14\) so all you're trying to solve is \(12x - 14 > 0\), which becomes \(x > \frac76\).

[dhaider]

Hey everyone, I need help with a probability question from HSC course.

Q) A certain type of bird has a probability of 1/12 of hatching a bird with white feathers. If a bird lays three eggs, find the probability of hatching:

i) exactly one bird with white feathers
iI) at least one bird with white feathers.

Any help or or solution will be appreciated guys. Thanks

[darkz]

Hey everyone, I need help with a probability question from HSC course.

Q) A certain type of bird has a probability of 1/12 of hatching a bird with white feathers. If a bird lays three eggs, find the probability of hatching:

i) exactly one bird with white feathers
iI) at least one bird with white feathers.

Any help or or solution will be appreciated guys. Thanks

Try using binomial probability i.e.
Let X be the number of birds with white feathers
Then, for i) Pr(X=1)
Then, for ii) Pr(X≥1)

(Success; p=1/12)
(Sample size; n = 3)