Indices / differentiation

f(x) = (3-4√x)² / √x Show that f(x).. equals:

9x^-1/2 + Ax1/2 + B where A and B are constants to be found.

Not quite sure how t0 get that - since I got something completely different after expanding brackets etc...

9 - 12 sprtx - 12 sprtx +16x / sprt x

16x - 24sprtx +9 / sprt x

Thats not going to look like how thay want it.... but not really sure what to do and the mark scheme doesnt give much away....

Reply 1

DrNippleson

4

Well, (3-4√x)² expands to 9 - 24√x +16x and if you put that all over √x you get 9x^-1/2 (1/x = -1 so 1/√x = -1/2), -24 (-24√x/√x) = -24) and +16x^1/2 (16x/√x = 16x^1/2 ---- x^1 - x^1/2 = x^1/2) so if I've differentiatied it correctly, A would be 16 and B would be -24

Reply 2

thefatone

6

Original post by christinajane

f(x) = (3-4√x)² / √x Show that f(x).. equals:

9x^-1/2 + Ax1/2 + B where A and B are constants to be found.

Not quite sure how t0 get that - since I got something completely different after expanding brackets etc...

9 - 12 sprtx - 12 sprtx +16x / sprt x

16x - 24sprtx +9 / sprt x

Thats not going to look like how thay want it.... but not really sure what to do and the mark scheme doesnt give much away....

f\left(x\right)= \dfrac{\left( 3-4 \sqrt x \right)^2}{\sqrt x}

\dfrac{9-12 \sqrt x - 12\sqrt x + \left( 4\times 4\right)\times \left(\sqrt x \times \sqrt x \right)}{\sqrt x}

remember that

\sqrt x = x^{\frac {1}{2}}

\dfrac {9-24\sqrt x +16x}{\sqrt x}

Spoiler

(edited 8 years ago)

Reply 3

christinajane

OP

4

Original post by DrNippleson

Well, (3-4√x)² expands to 9 - 24√x +16x and if you put that all over √x you get 9x^-1/2 (1/x = -1 so 1/√x = -1/2), -24 (-24√x/√x) = -24) and +16x^1/2 (16x/√x = 16x^1/2 ---- x^1 - x^1/2 = x^1/2) so if I've differentiatied it correctly, A would be 16 and B would be -24

Ahh I htink I get it now - each term is divided by that sprt x - because I was just bringing the sprt x up at the very end of the equation.

This revising malarky is helping me out big time! haha forgetting little things like that!

Reply 4

DrNippleson

4

Original post by christinajane

Ahh I htink I get it now - each term is divided by that sprt x - because I was just bringing the sprt x up at the very end of the equation.

This revising malarky is helping me out big time! haha forgetting little things like that!

I've done the same in my mocks - just got to remember that divide is minus :smile:

glad I could help

Reply 5

Zacken

22

Original post by christinajane

Ahh I htink I get it now - each term is divided by that sprt x - because I was just bringing the sprt x up at the very end of the equation.

This revising malarky is helping me out big time! haha forgetting little things like that!

Pretty much, all you're doing is that:

Unparseable latex formula:

\displaystyle [br]\begin{equation*}\frac{a+b+c+d}{e} = \frac{a}{e} + \frac{b}{e} + \frac{c}{e} + \frac{d}{e}\end{equation*}

Etc... - obviously using the fact that

\frac{a^n}{a^m} = a^{n-m}

and

\frac{1}{\sqrt{x}} = x^{-1/2}

.

Indices / differentiation

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