Rom

$f(x) \text{ will be continuous if the each of the individual parts is continuous }\\ \text{and the value of those parts agree at }0 \\ \text{each part is composed of continuous functions so we just have to check values at 0}\\ f(0^-)= a \tan^{-1}(0) + b = b \\ f(0^+) = \lim \limits_{x \to 0}~\dfrac{\ln(1+\sin(x))}{x}\\ \text{Using L'Hopital's rule we take the derivatives of numerator and denominator and look at the ratio} \\ \left .\dfrac{d}{dx} \ln(1+\sin(x)) = \dfrac{\cos(x)}{1+\sin(x)} \right|_{x=0} = 1\\ \dfrac{d}{dx}=1 \text{ and thus the limit is }\\ f(0^+) = \lim \limits_{x \to 0}~\dfrac{\ln(1+\sin(x))}{x} = \dfrac 1 1 = 1 \\ \text{and thus }b=1$

$\text{The value of the derivative at 0 must also agree from both sides} \\ \text{I'm going to let you work this bit out}\\ \text{Take the derivatives of both terms of f(x) }\\ \text{one can be immediately evaluated at 0, the other you'll have to take the limit as }x\to 0 \\ \text{set the value }a \text{ such that the two values agree}$

sigh... you really can't figure these out?

15 decreased by the product of a number x and 4.

15-4x

You eat five slices of bread. Your friend eats two slices fewer than you eat. Write and expression that describes the number of slices your friend eats.

5-2=3 

looks to be 1 cubic centimeter at 20 degC

$f\left(\dfrac 1 x\right)+\dfrac 1 x f(-x) = 2x ~;\text{now let }x \to -\dfrac 1 x \\ f(-x) -x f\left(\dfrac 1 x\right) = -\dfrac 2 x~; \text{and multiply by }\dfrac 1 x \\ \dfrac 1 x f(-x) - f\left(\dfrac 1 x\right) = -\dfrac{2}{x^2}$

$\text{now add this to the original equation }\\ \dfrac 2 x f(-x) = 2x - \dfrac{2}{x^2} = \dfrac{2x^3-2}{x^2} \\ f(-x) = \dfrac{x^3-1}{x} = x^2 - \dfrac 1 x\\ f(x) = x^2 + \dfrac 1 x$

$f(2) = 2^2 +\dfrac 1 2 = \dfrac 9 2$

$p(x)=x^4+2x^3-x^2+ax+b = (q(x))^2,~\text{for some polynomial }q(x)$

$\text{we know q(x) will be of order 2 so write it as }\\ q(x)=q_2 x^2 + q_1 x + q_0 \\ (q(x))^2 = q_2^2 x^4+2 q_1 q_2 x^3+\left(q_1^2+2 q_0 q_2\right) x^2+2 q_0 q_1 x+q_0^2$

$\text{and we have equations }\\ q_2^2 = 1\\ 2q_1q_2 = 2\\ (q_1^2+2q_0q_2)=-1\\ 2q_0q_1=a\\ q_0^2=b$

$\text{clearly }q_2 = \pm 1\\ \text{suppose }q_2=1 \\ 2q_1 (1)=2,~q_1=1\\ (1+2q_0(1))=-1 \\ q_0=-1 \\ b=q_0^2 = 1 \\ a=2q_0 = 2$

$\text{now suppose }q_2=-1 \\ 2q_1(-1) = 2,~q_1=-1 \\ (1+2q_0(-1))=-1,~q_0=1 \\ b=q_0^2 = 1 \\ a = 2(-1)(-1) = 2$

$\text{so in both cases }a=2,~b=1 \\ \text{and }a+b = 3$

I kind of suspect there is a simpler way to solve this.

$f(x)+2f(-x)=\sin(x)\\ f(-x)+2f(x) = \sin(-x) = -\sin(x) \\ \text{add these two equations} \\ 3(f(x)+f(-x)) = 0\\ f(-x)=-f(x)\\ f(x) -2f(x) = \sin(x) \\ -f(x) = \sin(x)\\ f(x) = -\sin(x) \\ f\left(\dfrac n 2\right) = -\sin\left(\dfrac n 2\right)$

you can see the max the graph attains is 2 and thus a=2

you can see that the sine wave as a period of 4pi and thus the frequency, b=1/2

$\dfrac{x}{3+7} = \dfrac{x}{10}$

$\dfrac{200}{y}-17$

deleted

The most straightforward way to do this is to just do the division.

It's hard to show the process of long division on here.

$\dfrac{1}{(1+x)^2}=\dfrac{1}{1+2x+x^2}$

and just set this up as a usual long division, if you go through it a few steps you'll see it's

$\dfrac{1}{1+2x+x^2} = 1 - 2x + 3x^2 - 4x^3 + 5x^4 \dots + (-1)^k (k+1)x^k,~\forall k \geq 0$

hrm.. there might be another way to approach it that's pretty straightforward

$\sum \limits_{k=0}^\infty ~x^k = \dfrac{1}{1-x}$

$\dfrac{1}{1+x}=\dfrac{1}{1-(-x)}$

$\dfrac{1}{1-(-x)} = \sum \limits_{k=0}^\infty ~(-1)^k x^k$

$\dfrac{d}{dx}~\dfrac{1}{1+x} = - \dfrac{1}{(1+x)^2}$

$\dfrac{1}{(1+x)^2} = -\dfrac{d}{dx} \left(\sum \limits_{k=0}^\infty~(-1)^k x^k\right) = \\ -\left(\sum \limits_{k=1}^\infty~(-1)^k k x^{k-1}\right) = \sum \limits_{k=0}^\infty~(-1)^k(k+1)x^k$