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\(f(x)=4\sqrt{x}+\sqrt[4]{x}\\ f(x)=4*x^{0.5}+x^{0.25}\\ f'(x)=2x^{-0.5}+0.25x^{-0.75}\\ f'(16)=2*16^{-0.5}+0.25*16^{-0.75}\\ f'(16)=2*4^{-1}+0.25*16^{3/4*-1}\\ f'(16)=2*4^{-1}+0.25*2^{3*-1}\\ f'(16)=2*4^{-1}+0.25*8^{-1}\\ f'(16)=\frac{2}{4}+\frac{0.25}{8}\\ f'(16)=\frac{16}{32}+\frac{1}{32}\\ f'(16)=\frac{17}{32}\\ \)

The gradient for part b is NOT 5/2

However, it is positive so the curve is increasing.

I really did not need to do all that. As soon as the curve was differentiated it was clear that the for all positive values of x the curve would be increasing.   I mean the first derivative would be positive.

thank you very much

Just tried this one for myself.....our answers agree...so....probably correct...!!!

y =  6x / [ x^3 + 4]

The domain will only be affected by any x that makes the denominator = 0

So....set the denominator to 0 to find this value

x^3 + 4  = 0

x^3 =  -4     take the cube root of both sides

x =  ∛-4  =  - ∛4

So....the domain is

(-infinity,  -∛ 4 ) U  ( - ∛4, infinity )    

Here's a graph : https://www.desmos.com/calculator/qih0ucpatr

Make sure that x^3 + 4 is not 0. Dividing by 0 is undefined. This means that \(x^3 \ne -4 \Rightarrow x \ne ^3\sqrt{-4} \Rightarrow x \ne 2i\).

This value is not real, so we do not need to consider it. The domain of \(f(x) = \dfrac{6x}{x^3+4}\) is \(x \in (-\infty, \infty)\).

- PM

y = 2tan ( θ) + 1

Verticle Stretch - the "2" indicates that this graph is stretched by a factor of 2 from the normal tangent graph

Period  -  same period as the normal tangent graph......pi rads

Phase Shift   -  there is no phase shift

Verticle Shift  - the "1"  means it is shifted up one unit

Y-intercept....let   θ  =  0    and we have

y = 2 tan (0) + 1

y =   2(0) + 1

y = 1   ⇒   y intercept

Here's the graph : https://www.desmos.com/calculator/v5o9vci8rf

Where can I get the θ (theta) symbol that looks like an O with a / through it? (The one that jbouyer used)

In \(\LaTeX\), it is \(θ\).

\(f(x)=\frac{1}{3x^2}=\frac{1}{3}x^{-2}\\ f'(x) =\frac{-2}{3}x^{-3}\\ f'(-1) =\frac{-2}{3}\times (-1)^{-3}\\ f'(-1) =\frac{-2}{3*(-1)^{3}}\\ f'(-1)=\frac{-2}{3*-1}\\ f'(-1)=\frac{2}{3}\\ \)

Since the gradient at x=-1 is positive, the graph is increasing.

You can also see this from PM's graph. you can see that the gradient of the tangent to the curve is positive when x=-1

If you want to check, b = 2, P = 0, and Q = 0. 

\((0\cdot 2^2) = 0\cdot 0\cdot2\) simplifies to \(0=0\) because anything multiplied by 0 is 0.

- PM

y = - 2sin (3θ) -pi/3   

Amplitude =  l -2 l   =  2

Period =  the "3" indicates that there are 3 periods in 2pi....thus each period is 2pi/3 rads long

Verticle Shift  =  -pi/3 units down 

Y- intercept.....to find this, set  θ to  0   and we have :

y = -2 sin (3 * 0)  - pi / 3

y =  -2 sin (0) - pi / 3

y = 0 - pi/3

y =  - pi/ 3  ⇒   y intercept

Here's the graph : https://www.desmos.com/calculator/er5ysgihkp

Let's try to express \((Pb^2) = QPb\) in terms of b. Since Q = P/2 and P = b - 2, \(((b - 2) \cdot b^2) = \dfrac{b-2}{2} (b - 2) b \Rightarrow b^3 - 2b^2 = \dfrac{1}{2}b^3 -2b^2 + 2b\). Combining like terms, we have \(\dfrac{1}{2}b^3 = 2b \Rightarrow b^3 = 4b\). Dividing both sides by b, we have \(b^2 = 4 \Rightarrow b = \pm 2\), which in this case, changes to b = 2. We know that P is two less than b (P = b - 2), so P = 2-2 = 0. 

\(\boxed{P = 0}\).

 - PM

Yes that is right, the Kg cancel out leaving you just with 1metre.    

1m^1=m

if you want to do it just with the indices rules you have

\(a=\frac{1Kg\times m^{-1}}{1Kg\times m^{-2}}\\ a=1m^{-1--2}\\ a=1m\)

ahhh ok thank you very much

So there you have it... the height of the water decreases around \(t = 2.25\).

Hope this helps,

 - PM

unfortunately the reason why i typed it in to desmos is i thought there would be roots to help me solve it but the range is a bit far for me haha 

Take the derivative

h ' =   12t^2 - 59t + 72

Set this to 0    and set to 0

12t^2 - 59t + 72 =  0

(4t -9) (3t - 8) = 0

Setting both factors to 0 and solving for x gives

t = 9/4          t  = 8/3

Take the second derivative

24t - 59

Plugging   9/4 into this gives   - 4....so....we have a max at  t = 9/4 sec = 2,25 sec

Plugging  8/3 into this  gives  5....so....we have a min at t = 8/3 sec = 2.66 sec

So...the water height is decreasing from  2.25 sec to 2.66 sec

Heres the graph : https://www.desmos.com/calculator/ne2dnm3tks

YEEEEEET.......This does look like a straight line if you don't "zoom in"....very subtle....!!!!

That is where the graph caught you off guard! The graph is actually not a straight line. Zoom out, and you will see that the graph curves like an "N" at around y = 58! 

- PM

a and b should both be increasing and the numbers i showed are to explain why...

my question is how to get to those numbers...