Mathematics functins and relations.... (1 Viewer)

[Mdzabakly]

Umm any help and i will be greatfull!!!
( also can i get full working out cuz im one of those kids who needs it =-= sorry)


1.The curve y=ax^3+bx passes through the point (1,7). The tangent at this point is parallel to the line y=2x-6
Find the values of a and b


2.Show that y= -sqr(36-x^2) is an even function

3. ( from ii) )

 

[Mdzabakly]

BUMP plz help and soz for extra bump

 

[Siddy123]

is you are doing gen maths ?

 

[nexus_lad_69]

1. dy/dx = 3ax^2 + b

f'(1) = 3ax^2 + b = 3a(1)^2 + b
mT = 3a^2 + b
= 3a + b

mT = m1 for ||
m1 of y = 2x-6 is 2.
mT = 3a + b, m1 = 2

3a + b = 2... eq1
b = 2-3a .... eq3
y = ax^3 + bx --> 7 = a(1)^3 + b(1) --> a + b = 7 ..... eq2
sub 3 into 2
a + (2 - 3a) = 7
-2a = 5
a = -5/2
b = 7 - a = 7 + 5/2 = 19/2

Therefore, a = -5/2 & b = 19/2

2. Just show that f(-x) = f(x). Therefore, even function.

 

[nexus_lad_69]

3. i. f(0) = So which domain includes 0? The second one does.
So therefore, f(0) refers to 9 - x^2
f(0) = 9 - (0)^2 = 9
ii. I'm sure you can sketch that. Just follow the domains.
iii. Look at your sketch and find the highest and lowest y-values to determine the range which the piece-meal function exists in.

 

[Mdzabakly]

is you are doing gen maths ?

umm no mathmatics XD

 

[Mdzabakly]

umm yea still dont know how to do Show that y= -sqr(36-x^2) is an even function ( although i knew about the f(-x)=f(x) part idk how to do it...

 

[Timske]

umm yea still dont know how to do Show that y= -sqr(36-x^2) is an even function ( although i knew about the f(-x)=f(x) part idk how to do it...

<a href="http://www.codecogs.com/eqnedit.php?latex=f(x) = -\sqrt{36 -x^2}\\\\ f(-x)= -\sqrt{36 -(-x)^2} \\\\\ ~~~~~~~~~=-\sqrt{36-x^2} \\\\ ~~~~~~~~~=f(x) \\\\ f(-x) = f(x) ~,~\therefore even function" target="_blank"><img src="http://latex.codecogs.com/gif.latex?f(x) = -\sqrt{36 -x^2}\\\\ f(-x)= -\sqrt{36 -(-x)^2} \\\\\ ~~~~~~~~~=-\sqrt{36-x^2} \\\\ ~~~~~~~~~=f(x) \\\\ f(-x) = f(x) ~,~\therefore even function" title="f(x) = -\sqrt{36 -x^2}\\\\ f(-x)= -\sqrt{36 -(-x)^2} \\\\\ ~~~~~~~~~=-\sqrt{36-x^2} \\\\ ~~~~~~~~~=f(x) \\\\ f(-x) = f(x) ~,~\therefore even function" /></a>

 

[Mdzabakly]

<a href="http://www.codecogs.com/eqnedit.php?latex=f(x) = -\sqrt{36 -x^2}\\\\ f(-x)= -\sqrt{36 -(-x)^2} \\\\\ ~~~~~~~~~=-\sqrt{36-x^2} \\\\ ~~~~~~~~~=f(x) \\\\ f(-x) = f(x) ~,~\therefore even function" target="_blank"><img src="http://latex.codecogs.com/gif.latex?f(x) = -\sqrt{36 -x^2}\\\\ f(-x)= -\sqrt{36 -(-x)^2} \\\\\ ~~~~~~~~~=-\sqrt{36-x^2} \\\\ ~~~~~~~~~=f(x) \\\\ f(-x) = f(x) ~,~\therefore even function" title="f(x) = -\sqrt{36 -x^2}\\\\ f(-x)= -\sqrt{36 -(-x)^2} \\\\\ ~~~~~~~~~=-\sqrt{36-x^2} \\\\ ~~~~~~~~~=f(x) \\\\ f(-x) = f(x) ~,~\therefore even function" /></a>

Umm wouldnt ( on the second step) the -(-x)^2 become +x^2?
(this is where i got lost in the first place ...)

 

[Timske]

-(-x)^2 = -(+x^2) = -x^2

 

[girlworld_club]

just wondering how you got from the 3rd to fourth line.

 

[starshine02]

F(x) is proven in first line. F(-x) proven in third and these two are equal so that makes F(x)=F(-x)

 

[Leffife]

just wondering how you got from the 3rd to fourth line.

If you look at the 3rd line it says: - squareroot (36 - x²) and this is exactly the same as f(x) which is how they got from step 3 to 4.

 

[Leffife]

Umm wouldnt ( on the second step) the -(-x)^2 become +x^2?
(this is where i got lost in the first place ...)

uhh... no. - (-x)^2 will be -x^2
If you still don't get it think of it like this - (-x.-x). In side the brackets when done - (x^2) = -x^2

 

[Leffife]

@ Nexus_lad_69

Hi Devil Jin