      #### Inverse arccos

Let $f(x)= -3\pi +2 \arccos\left( e^{3x+1}\right)$ with range $D'_f = \left[ -3\pi, -2\pi\right[$. Consider that $f^{-1}(x)$ is the analytic expression of the inverse function of $f$. State whether the following statements are true (T) or false (F).

# Complements of differential calculus in real numbers

Training
13      #### Inequality arccos

Let $f(x)= -3\pi+2\arccos(3x+1)$ of domain $D_f=\left[-\dfrac{2}{3}, 0\right]$. The solution of inequality $f(x) > \arccos\left( -\dfrac{1}{2}\right) - 3 \arccos\left(-1\right)$ is:

# Complements of differential calculus in real numbers

Training
13      #### Equation arccos

Consider the functions $f(x)= 5 \arccos\left(2x\right)$ with domain $D_f=\left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$ and $g(x)= \arccos\left(\dfrac{x}{2}\right)-3\pi$ with domain $D_g=\left[-2, 2\right]$. The solution of the equation $f(x) = 5 \ g(x)+15\pi$ is:

# Complements of differential calculus in real numbers

Training
13      #### Equation arccos

Let $f(x)= -3\pi +2 \arccos\left( e^{3x+1}\right)$ with domain $D_f=\left]-\infty, -\dfrac{1}{3}\right]$. $f^{-1}(x)= -\dfrac{1}{3} + \dfrac{1}{3} \ln\left( \cos\left( \dfrac{x+3\pi}{2}\right)\right)$ is the analytic expression of the inverse function with domain $D_{f^{-1}}=\left[ -3\pi, -2\pi\right[$. The solution of the equation $f\left(\dfrac{x-1}{3}\right) -f\left(-\dfrac{1}{3}\right) -\arccos(0)=\sec\left( \arccos\left(\dfrac{1}{3}\right) \right) +\dfrac{1}{f^{-1}(-3\pi)}$ is:

# Complements of differential calculus in real numbers

Training
13      #### Derivation arccos

Let $f$ and $g$ be two differentiable functions in their domain, such that: $f(x)= -3\pi +2 \arccos(3x+1)$, $g(x)=\sin(f(x))$ and $D_f=\left[-\dfrac{2}{3}, 0\right]$. Considering that $g'(x)$ is the first derivative of function $g(x)$, then $g'\left(-\dfrac{1}{3}\right)$ is:

# Complements of differential calculus in real numbers

Training
13      #### Derivation arccos

Consider $g(x)= \arccos\left(\dfrac{x}{2}\right)-3\pi$ with domain $D_g=[-2, 2]$. The first derivative of the function $g(x)$ is given by $g'\left(x\right)=\dfrac{b}{\sqrt{a-x^2}}$. Then, the values of $a$ and $b$ are, respectively:

# Complements of differential calculus in real numbers

Training
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