Lighten up. No disrespect to Sammy intended but I do believe this is why people have such a terrible perception of math-oriented people, and professors too for that matter.
Life is too short to sweat the little things Mark and ill be damned if I am going to sit on here and be lectured by some...
An extremely poorly taught advanced functions class, that throws its students two small examples, slams them with 35 homework questions that go 6 steps further than taught and do me no good in terms of actually understanding material. I have solved the question. Thank you.
New to composite functions here. Lesson has been vague and unhelpful.. again. Here is what I've worked on so far but not sure on the last equation in particular, or that I have done my multiplication properly when working with a squared set of brackets, multiplied by an number.. (b and c)
Any...
Sorry for the readability. I don't understand how you could factor the x's out in the middle of the equation. Should I just divide both sides by -1 to get rid of that?
Unfortunately I am not supposed to be using derivatives. The only examples I have involve setting the equations equal and solving out to ax^2+bx+c=0, and using the discriminant of the quadratic formula (listed as b^2-4ac =0).
Can someone try to tackle this question for me? I am lost.
Having some trouble with this..
Need to find equation of a line with a slope of -1 that is tangent to the curve y=1/(x-1).
So, rearranging slope formula as y=-1x+k and setting the equations equal,
y=-1x+k=1/(x-1)
y=(-1x+k)(x-1)=1
Here is where my multiplication is either totally wrong or I am...
Hi All,
Having a tough time with this one and I'm not sure why.
Need to state amplitude, period and phase shift of f(x)=3cos2[x-(π/4)]+1.
Amplitude being 3, period being 2π/2=π and phase shifted (π/4) to the right.
Midline would also be at y=1
Good so far?
Right, so I know that 1/4 phase...
HALLELUJAH! Sir you are one brilliant individual. I could not figure out for the life of me why there was no material in the text on expanding out equations like this, and much less for two marks. CHEERS mate.
Im not sure if I am on the right track here or not..
f(x)=2x(x-3)^3(x+1)(4x-2)
=2x(x-3)^3(4x^2-2x-2+4x)
=2x^(2*3)-6x^3(8x^2-4x-4+8x)
Did I solve this out ok? If this were the case would 6 not be the degree and 2x^3 the dominant figure?
So you're saying to multiply (x+1)(4x-2) and then by the first factor? I can expand in general but not sure what to do with the 2x(x-3)^3 bit.
Multiplying the (x+1)(4x-2) you would obviously get 4x^2+2x-2..so how do you proceed from there?