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

Mdzabakly

Member
Joined
Apr 16, 2012
Messages
458
Gender
Male
HSC
2013
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. Untitled.jpg ( from ii) )
 

nexus_lad_69

Sikkest Lad in Livo
Joined
Jul 15, 2012
Messages
55
Location
Hektik Livo Hideout
Gender
Male
HSC
2009
Uni Grad
2015
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.
 
Last edited:

nexus_lad_69

Sikkest Lad in Livo
Joined
Jul 15, 2012
Messages
55
Location
Hektik Livo Hideout
Gender
Male
HSC
2009
Uni Grad
2015
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

Member
Joined
Apr 16, 2012
Messages
458
Gender
Male
HSC
2013
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

Sequential
Joined
Nov 23, 2011
Messages
794
Gender
Male
HSC
2012
Uni Grad
2016
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

Member
Joined
Apr 16, 2012
Messages
458
Gender
Male
HSC
2013
<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 ...)
 

Leffife

A lover is a best friend
Joined
May 10, 2012
Messages
578
Location
Heaven
Gender
Male
HSC
N/A
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
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top