Find the sum of the series. sum_(n=1)^(infty) 7/(4^n)

Express the point with cartesian coordinates Q(3,-1) in polar form

Find the domain of the vector-valued function. (Enter your answer using interval notation.)
${\displaystyle r\left(t\right)=sqr(9-t^2)i+t^{2}j-5tk}$

Consider the following convergent series.
a. Find an upper bound for the remainder in terms of n.
b. Find how many terms are needed to ensure that the remainder is less than ${10}^{-3}$.
c. Find lower and upper bounds (ln and Un, respectively) on the exact value of the series.
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{3^k}$

Tips for integrating $\int \frac{dx}{1+\cos (x)}$

Describe the motion of a particle when the tangential component of acceleration is 0.

What is the convergence rate of the tail of the series
${\displaystyle a_n=\sum _{k>n}\text{exp}(-pk)k{\left(\log k\right)}^2}$

Intermediate value theorem: show there exists $c\in [0,2/3]$ such that $f(c+\frac{1}{3})=f(c)$
Let $f:[0,1]\to [0,1]$ be continuous and $f(0)=f(1)$
This is pretty straight forward and the other ones ive done it was easier to find why IVT applies, namely find why it changes signs.
Letting $g(x)=f(x+\frac{1}{3})-f(x)$ then this is only defined for the interval $[0,\frac{2}{3}]$ b/c not sure what $g(1)=f(\frac{4}{3})-f(1)$ may be.
Now, $g(0)=f(\frac{1}{3})-f(0)$
and $g(\frac{2}{3})=f(1)-f(\frac{2}{3})=f(0)-f(\frac{2}{3})$.
I can't (at least I don't see why I can) conclude this is equal to $-g(0)$ which has been the case in other exercises, which is what I used to apply the IVT and get the wanted result.