question about derivatives (1 Viewer)

Nailgun · Jul 10, 2016

I was just doing a past paper, and I noticed that in the solutions they found the max-min of a function by using the square of the function?

Like, you needed to find the value of t which maximises f(t)

but then instead of finding df(t)/dt, they used d(f(t))^2/dt

can someone explain the rules behind this

they got the same answer as I did, but their way was way easier to differentiate

InteGrand · Jul 10, 2016

Nailgun said:
I was just doing a past paper, and I noticed that in the solutions they found the max-min of a function by using the square of the function?

Like, you needed to find the value of t which maximises f(t)

but then instead of finding df(t)/dt, they used d(f(t))^2/dt

can someone explain the rules behind this

they got the same answer as I did, but their way was way easier to differentiate

$\noindent If $f(t)$ is something that it always non-negative (e.g. a distance), then $t^*$ maximises (respectively minimises) $f$ if and only if $t^*$ maximises (respectively minimises) $f^2$, because the square function is increasing on $\left[0, \infty \right)$.$

Nailgun · Jul 10, 2016

InteGrand said:
$\noindent If $f(t)$ is something that it always non-negative (e.g. a distance), then $t^*$ maximises (respectively minimises) $f$ if and only if $t^*$ maximises (respectively minimises) $f^2$, because the square function is increasing on $\left[0, \infty \right)$.$

So if the function is always positive, all values that maximise f(t) will also maximise [f(t)]^2?

InteGrand · Jul 10, 2016

$\noindent We can generalise this as follows. Let $f$ be a (real-valued) function defined on a subset $X$ of $\mathbb{R}$. Let $\phi$ be a (real-valued) function defined on the range of $f$. If $\phi$ is an increasing function (on range of $f$), then $t^*$ maximises $f(t)$ (respectively minimises) $\phi \left(f(t)\right)$ iff it maximises (respectively minimises) $\phi\left(f(t)\right)$.$

$\noindent If $\phi$ is instead a decreasing function, then we switch mins and maxes around, i.e. a $t$ that minimises $f(t)$ \emph{maximises} $\phi \left(f(t)\right)$, etc.$

The above ideas are quite useful for some optimisation problems in HSC.

InteGrand · Jul 10, 2016

Nailgun said:
So if the function is always positive, all values that maximise f(t) will also maximise [f(t)]^2?

$\noindent Yes (and vice versa), so finding the $t$ that optimises $f$ becomes equivalent to finding the $t$ that optimises $\left(f(t)\right)^2$. Sometimes optimising the square is easier (usually the case for distance problems since it means we don't need to worry about a square root, which is messier to differentiate).$

Nailgun · Jul 10, 2016

A BoS classic

"You must spread some Reputation around before giving it to InteGrand again."

Thank you btw

question about derivatives (1 Viewer)

Nailgun

Cole World

InteGrand

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Nailgun

Cole World

InteGrand

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InteGrand

Well-Known Member

Nailgun

Cole World

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