Question

Find all relative extrema of the following functions and determine if they are relative maximum or relative minima.

Answer 1

\[f(x)=\frac{ t^{4} }{ (t-1)^2 }\]

Answer 2

Were you able to find f'(x)?

Answer 3

no, I'm not sure how to derive it

Answer 4

Do you know how to take a derivative?

Answer 5

kind of. would i need to use the quotient rule?

Answer 6

Yes quotient rule :)

Answer 7

Sure. Quotient rule will work fine :)

Answer 8

Take your first derivative. Set this equal to zero, and find all the points that make this true. These are candidates for your extrema.

The second derivative will tell you which kind of extrema these points are, but we'll get to that in a bit.

Answer 9

is it possible for someone to work out the steps of it though? the square on the bottom confuses me

Answer 10

\[f(x) = \frac{high(x)}{low(x)}\]

\[f'(x) = \frac{low * d(high) - high*d(low)}{low^2}\]

Answer 11

For the derivative of the "low" function, you'll need to use the chain rule. For example:
\[low(x) = (x^2-1)^2\]
\[\frac{d(low)}{dx} = 2(x^2-1)*2x = 4x(x^2-1)\]

Answer 12

Another way to think of the chain rule is to use a substitution.

If f(x) = (x^2 - 1)^2, let u = x^2-1.

Then, du/dx = 2x, and f(u) = u^2, so df/du = 2u

Then, we can find df/dx by:
\[\frac{df}{dx} = \frac{d f(u)}{du}*\frac{du}{dx}\]
Plug in what we know, and we're done!

Answer 13

its (t-1)^2 though not (x^2-1)^2

Answer 14

I know. I want you to do that and tell me what you get :)

Answer 15

Your function is a rationale function, that is a quotient a two polynomials.

Let \[u(t)=t^4\] and \[v(t)=(t-1)^²\].

Hoping you can derive polyomials, this rresults in \[u'(t)=4t^3\] and \[v'(t)=2(t-1)\].

Now, as \[f(t)=\frac{ u(t) }{ v(t) }\], you must use the quotient rule for the derivatives :

\[f'(t)=\frac{ u'(t)v(t)-u(t)v'(t) }{ v(t)^2 }\].

Finally, you plug the formulas for u, u', v and v' into f' formula, and symplify the result using algebra.

Try it. :)

Answer 16

so i got (using substitution) \[f(x)=\frac{ u^{2}*4t^3-t^4*2u }{ u^4 }\]

Answer 17

Hmm, I wasn't clear on the substitution. You have to substitute back in for u before using it in other steps.

\[f(t) = \frac{t^4}{(t-1)^2}\]
\[g(t) = t^4, g'(t) = 4t^3\]
\[h(t) = (t-1)^2,h'(t) = 2(t-1)\]

For clarity on the substitution, we'd use it for the (t-1)^2 term:
\[h(t) = (t-1)^2 = u^2\]
\[u = t-1, du = 1\]
\[h'(t) = h'(u)*du = 2u*1 = 2(t-1)\]

Now, use what we've found for g, g', h and h' to find the overall derivative.

Answer 18

Be careful to notation.

\[f\] is the function of the single variable \[t\]. (not x here)

\[f'\] denotes the derivative of the function \[f\]

When using substitution, you have to make a clear distinction of the fun000000ctions you are using.

Answer 19

I recommend you to review what is a function and what is "composition of functions" before computing derivatives.

Answer 20

is it \[f'(x)= \frac{ 2(x-2) x^3 }{ (x-1)^3 }\]

Answer 21

Your original function was given in terms of t, not x,
but yes that derivative and simplification looks correct! :)

Answer 22

ohh ok so I set it equal to 0 and then I get 0 and 2, which are my critical points

Answer 23

Good good good.

Answer 24

how do i determine if they are relative max/min?

Answer 25

Umm let's run some test points into the derivative.