Goodwin Lang
02/02/2024 · High School
1.2.1. \( F(x)=\frac{x^{4}-5 x^{3}+\sqrt{x}}{x^{2}} \) 1.2.2. \( H(t)=\left[\sin ^{-1}(3 t)+2^{\ln (5)}\right]^{9} \) 1.2.3. \( G(x)=e^{\cot (2 x)} \tan ^{-1}\left(x^{2}\right) \)
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Step-by-step Solution
Let's analyze and simplify each of the given functions step by step.
### 1.2.1. \( F(x)=\frac{x^{4}-5 x^{3}+\sqrt{x}}{x^{2}} \)
To simplify \( F(x) \), we can divide each term in the numerator by \( x^2 \):
\[
F(x) = \frac{x^4}{x^2} - \frac{5x^3}{x^2} + \frac{\sqrt{x}}{x^2}
\]
This simplifies to:
\[
F(x) = x^2 - 5x + \frac{1}{\sqrt{x}} \quad \text{(for } x > 0\text{)}
\]
### 1.2.2. \( H(t)=\left[\sin^{-1}(3t)+2^{\ln(5)}\right]^{9} \)
First, we simplify \( 2^{\ln(5)} \). Using the property of exponents and logarithms, we have:
\[
2^{\ln(5)} = e^{\ln(5) \ln(2)} = 5^{\ln(2)}
\]
However, for simplicity, we can leave it as \( 2^{\ln(5)} \). Thus, we can rewrite \( H(t) \):
\[
H(t) = \left[\sin^{-1}(3t) + 2^{\ln(5)}\right]^{9}
\]
This function is already in a simplified form, but we note that \( \sin^{-1}(3t) \) is defined for \( -\frac{1}{3} \leq t \leq \frac{1}{3} \).
### 1.2.3. \( G(x)=e^{\cot(2x)} \tan^{-1}(x^2) \)
The function \( G(x) \) is a product of two functions: \( e^{\cot(2x)} \) and \( \tan^{-1}(x^2) \). There is no further simplification possible without specific values for \( x \). Thus, we can express it as:
\[
G(x) = e^{\cot(2x)} \tan^{-1}(x^2)
\]
### Summary of Simplified Functions
1. \( F(x) = x^2 - 5x + \frac{1}{\sqrt{x}} \) for \( x > 0 \)
2. \( H(t) = \left[\sin^{-1}(3t) + 2^{\ln(5)}\right]^{9} \) for \( -\frac{1}{3} \leq t \leq \frac{1}{3} \)
3. \( G(x) = e^{\cot(2x)} \tan^{-1}(x^2) \)
These are the simplified forms of the given functions.
Quick Answer
1. \( F(x) = x^2 - 5x + \frac{1}{\sqrt{x}} \) for \( x > 0 \)
2. \( H(t) = \left[\sin^{-1}(3t) + 2^{\ln(5)}\right]^{9} \) for \( -\frac{1}{3} \leq t \leq \frac{1}{3} \)
3. \( G(x) = e^{\cot(2x)} \tan^{-1}(x^2) \)
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