Mathematics Question Thread

[alexnero7]

Anyone know how to do this?

[meerae]

Anyone know how to do this?

Hey alexnero7!
You can differentiate this by using the quotient rule, you would differentiate it twice to find first and second derivative.

Hope this helped!
meerae

[jamonwindeyer]

Anyone know how to do this?

Here's the first one!



Have a shot at the second one, it is really similar!

[abdulrahmanb8]

Use the quotient rule for differentiation

[shaynec19]

Can someone please help me to solve this definite integral.

[RuiAce]

Can someone please help me to solve this definite integral.

\begin{align*}\int_3^4 \frac{x^2+x+3}{3x^5}\,dx &= \frac13 \int_3^4 \left( \frac{x^2}{x^5}+\frac{x}{x^5}+\frac{3}{x^5} \right)dx\tag{split fraction up}\\ &= \frac13 \int_3^4 \left( x^{-3}+x^{-4}+3x^{-5}\right)dx \end{align*}
You should be able to take it from here.

[shaynec19]

Thanks Rui!

[alexnero7]

Hi can someone please help me solve this? For some reason I'm getting an inflexion point, but it should be getting a minimum. Thanks.

[RuiAce]

Hi can someone please help me solve this? For some reason I'm getting an inflexion point, but it should be getting a minimum. Thanks.

It is true that if we have a point of inflexion, the second derivative is equal to 0.
However, it is not always true that if the second derivative equals to 0, we have a point of inflexion.

For this reason, once we find \( \frac{d^2y}{dx^2} = 0 \) (which is true at your stationary point \(x=0\)), we must always test both sides of the equation to see if there is a concavity change. That forms the distinction between a horizontal point of inflexion and a turning point.

Here, testing a bit to the left, say \(x=-1\) we have \( \frac{d^2y}{dx^2} = 1 > 0\). Testing a bit to the right, say \(x = 1\) we have \( \frac{d^2y}{dx^2} = 1 > 0 \). Hence there is no concavity change, and thus we do not have a point of inflexion.

In fact, because the concavity remains concave up, we deduce it is a local minimum.

[alexnero7]

Thanks Rui

It is true that if we have a point of inflexion, the second derivative is equal to 0.
However, it is not always true that if the second derivative equals to 0, we have a point of inflexion.

For this reason, once we find \( \frac{d^2y}{dx^2} = 0 \) (which is true at your stationary point \(x=0\)), we must always test both sides of the equation to see if there is a concavity change. That forms the distinction between a horizontal point of inflexion and a turning point.

Here, testing a bit to the left, say \(x=-1\) we have \( \frac{d^2y}{dx^2} = 1 > 0\). Testing a bit to the right, say \(x = 1\) we have \( \frac{d^2y}{dx^2} = 1 > 0 \). Hence there is no concavity change, and thus we do not have a point of inflexion.

In fact, because the concavity remains concave up, we deduce it is a local minimum.

[alexnero7]

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[jamonwindeyer]

.

Hey! Curve sketches like this are a really common HSC question, so common that I actually wrote a guide on them! Read it here, it might be enough to get you through! Otherwise feel free to let us know which step you get stuck on

[emilyyyyyyy]

Hi all, can someone please tell me how to answer the question attached? I've completely forgotten how to
Thanks

[fun_jirachi]

Hey there!

All you have to do for this question is find the second derivative  Since you're given that x=2 and x=-1 are points of inflexion, you have that f''(2) and f''(-1)=0. Solve simultaneously to find a and b. Answers in the spoiler

Spoiler

a=3, b=-3

[alexnero7]

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.