quick easy maths question (1 Viewer)

[thecan]

ive had a mental blank

make x the subject...

y=3^-x

 

[Trev]

y=3^-x
log₃=-x
x=log₃(1/y)

 

[thecan]

hmm, to quote the original question was....

state the range of y=3^-x

i know i need to make x the subject, but i dont think i need to get into logs???

 

[webby234]

The range of the function is y > 0

You don't need to make x the subject, although that is one way to solve it. You can not take the log of a negative number and 1/0 is undefined.

Other ways are to draw it as a graph - exponential graph asymptoting at y = 0, x approaching infinity.

Or consider that the function is positive as x approaches negative infinity and positive as x approaches positive infinity. There are no stationary points and the function is defined for all real x, so it is therefore positive for all real x.

 

[SoulSearcher]

Exponential function, range is y > 0

 

[webby234]

Soulsearcher, doesn't work for y = 0.

 

[SoulSearcher]

lol, just realised that just then. All fixed now

 

[thecan]

thanks. stupid ext one assessment, need to redo the ones i got wrong. any good with trig

 

[thecan]

simplify....

cos^2θsin2θ+cos^4θ

and

find the exact value of the following.....

5cos30ºsin225º

 

[webby234]

5cos30ºsin225º

= 5rt(3)/2 * -1/rt(2)

= -5rt3/2rt2

= -5rt6/4

 

[webby234]

cos^2θsin2θ+cos^4θ

= cos²θ * 2cosθsinθ + cos⁴θ

= 2cos³θ sinθ + cos⁴θ

= cos³θ (2sinθ + cosθ)

Not sure if that's the answer they're looking for - its hard to tell when it just says simplify.

 

[thecan]

yeh sorry, theres been a mistake in the question

the question should read...

cos^2θsin^2θ+cos^4θ

notice that the sin is sin^2 not sin2
oops

to provide some insight into the answers a question like this could be simplified like so:

cotθsinθsecθ
= 1/tanθ * sinθ/1 * 1/cosθ

=1/tanθ * sinθ/cosθ

=1/tanθ * tanθ/1

=1

and what does rt mean?

 

[rama_v]

and what does rt mean?

square root

 

[webby234]

cos^2θsin^2θ+cos^4θ

= cos²θ (1 - cos²θ) + cos⁴θ

= cos²θ - cos⁴θ + cos⁴θ

= cos²θ