Melody

Thanks Chris,

I am not entirely comfortable with questions like this so it was really good for me to work through your answer :))

Thanks Alan :)

whats 4 small 8 time 4 indices

huh?????

\(4_8*4\)?????

Kule sounds a little harsh but I expect he means the same as I meant when I said.

Ask one at a time and LEARN from the answer so that you can try the next one on your own.

We don't mind helping with lots of questions so long as we feel you are actually learning from our answers.  

I would like another mathematician to read what I have put and tell me if they agree that it is entirely right and if it is not then tell me why.

True or False. If f is continuous on [2,4], and f1(c)= (f(4)-f(2))/2,  then c ∈ [2,4].

I think that you mean:

True or False. If f is continuous on [2,4], and f'(c)= (f(4)-f(2))/2,  then c ∈ [2,4].

(f(4)-f(2))/2 is the gradient of the line joining the points (2,f(2)) and (4,f(4))

so yes of course, there must be a point on f(x) between 2 and 4 where the gradient of the tangent is equal to that.

So 

yes 

There is at least one c  between [2,4] that will make that true.     There could be other c's elsewhere that make it true too.

If you draw a pic and think about it you might realize why I think the answer is obvious  (I hope I am entirely right)

I did a very similar one to this a moment ago

Hi Girldesi :)

If g(x)=x4-2, find g'(1) and use it to find an equation of the tangent line to the curve y=x^4-2 at the point (1,-1).

\(g(x)=x^4-2\\ g'(x)=4x^3\\ g'(1)=4*1^3\\ g'(1)=4\\ \)

The thing you must ALWAYS remember is the the first dirivative at a particular point IS the gradient of the tangent to the curve at that point.

so now you just have to find the line with gradient -4 through the point (1,-1)

You should be able to do that yourself but if you can't ask me to show you.

--------------------------------------------------------------------------------

Askers comment:  I dont know how to solve using the equation: f(x+h)-f(x)/h

I have already answered but now I iwill show you hos to find the gradient of the tangent from first principals

\(\begin{align} gradient&=\displaystyle\lim_{h \rightarrow 0}\;\frac{f(x+h)-f(x)}{h}\qquad where\;\;x=1\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{((1+h)^4-2)-(1-2)}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{((1+h)^4-1}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{h^4+4h^3+6h^2+4h+1-1}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{h^4+4h^3+6h^2+4h}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\frac{h(h^3+4h^2+6h+4)}{h}\\ &=\displaystyle\lim_{h \rightarrow 0}\;\: h^3+4h^2+6h+4\\ &=4 \end{align} \)

Again you still have to find the equation of the line through (1,-1) with a gradient of 4

If you need more explanation then please ask for it :))

Sawcut grass??  Is that grass that is cut with a saw??  How do you cut grass with a saw?   Hahaha  

I was so busy answereing I forgot to say hello.

Welcome back to the fourm songoflight :))

Say everyone one in your class does a test.

You order the papers from best to worst.  

The mode is the score that happens the most often,  there might be no mode or one mode or any number of modes.

The median is the score on the paper that is right in the middle of the pile.  OR if there are 2 middle papers than the median is the average of the 2 middle scores.

Do you know the Hey Didile Diddle nursery rhyme ?

Here is a statistical version of it.

Hey diddle diddle,

The median's the middle,

You add and divide for the mean,

The mode's the one you see the most and the range is inbetween     :))