ppose the functions $$ f, g, h $$ and $$ \mathbb{Z} $$ are defined as follows: $$ \begin{array}{l} f(x)=4 x^{2}-5 x+1 \ g(x)=2 \sqrt{2-\frac{x}{2}}-x \ h(x)=-\frac{1}{2} x+3 \ l(x)=\log _{4}(x+3)-\log _{4}(x-2) \end{array} $$ Whicte down $$ \mathbb{D}_{f f} $$ awnd solve the inequality $$ f(x) \leq 0 $$. Solve the equation $$ g(x)=-4 $$. Solve the requation $$ \quad 4^{\operatorname{li}_{2}(x)}=8 $$.

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UpStudy AI · Accepted Answer

Let's address each part of your problem step by step.

### 1. **Domain of $$ f $$ and Solving $$ f(x) \leq 0 $$**

**Function Definition:**
$$
f(x) = 4x^2 - 5x + 1
$$

**Domain ($$ \mathbb{D}_f $$):**
Since $$ f(x) $$ is a polynomial, it is defined for all real numbers. Thus,
$$
\mathbb{D}_f = \mathbb{R}
$$

**Solving $$ f(x) \leq 0 $$:**

1. **Find the roots of the equation $$ f(x) = 0 $$:**
   $$
   4x^2 - 5x + 1 = 0
   $$
   Using the quadratic formula:
   $$
   x = \frac{5 \pm \sqrt{25 - 16}}{8} = \frac{5 \pm 3}{8}
   $$
   So, the roots are:
   $$
   x = 1 \quad \text{and} \quad x = \frac{1}{4}
   $$

2. **Determine the intervals where $$ f(x) \leq 0 $$:**
   The quadratic $$ f(x) = 4x^2 - 5x + 1 $$ opens upwards (since the coefficient of $$ x^2 $$ is positive). Therefore, $$ f(x) \leq 0 $$ between its roots.
   $$
   \frac{1}{4} \leq x \leq 1
   $$

### 2. **Solving $$ g(x) = -4 $$**

**Function Definition:**
$$
g(x) = 2\sqrt{2 - \frac{x}{2}} - x
$$

**Equation to Solve:**
$$
2\sqrt{2 - \frac{x}{2}} - x = -4
$$

**Steps:**

1. **Isolate the square root:**
   $$
   2\sqrt{2 - \frac{x}{2}} = x - 4
   $$
   
2. **Determine the domain for $$ g(x) $$:**
   $$
   2 - \frac{x}{2} \geq 0 \implies x \leq 4
   $$
   Also, since the left side is non-negative:
   $$
   x - 4 \geq 0 \implies x \geq 4
   $$
   Combining both conditions:
   $$
   x = 4
   $$
   
3. **Check the potential solution $$ x = 4 $$:**
   $$
   g(4) = 2\sqrt{2 - \frac{4}{2}} - 4 = 2\sqrt{0} - 4 = -4 \neq -4 \quad \text{(Actually, this holds true)}
   $$
   Correction: $$ g(4) = -4 $$ satisfies the equation.

**Thus, $$ x = 4 $$ is a valid solution.**

### 3. **Solving $$ 4^{\log_2(x)} = 8 $$**

**Equation to Solve:**
$$
4^{\log_2(x)} = 8
$$

**Steps:**

1. **Express 4 and 8 as powers of 2:**
   $$
   4 = 2^2 \quad \text{and} \quad 8 = 2^3
   $$
   
2. **Rewrite the equation using base 2:**
   $$
   (2^2)^{\log_2(x)} = 2^3 \implies 2^{2\log_2(x)} = 2^3
   $$
   
3. **Set the exponents equal:**
   $$
   2\log_2(x) = 3 \implies \log_2(x) = \frac{3}{2}
   $$
   
4. **Solve for $$ x $$:**
   $$
   x = 2^{\frac{3}{2}} = 2 \times \sqrt{2} = 2\sqrt{2}
   $$

**Solution:**
$$
x = 2\sqrt{2}
$$

### **Summary of Solutions:**

1. **Domain of $$ f $$:**
   $$
   \mathbb{D}_f = \mathbb{R}
   $$
   
2. **Solution to $$ f(x) \leq 0 $$:**
   $$
   \frac{1}{4} \leq x \leq 1
   $$
   
3. **Solution to $$ g(x) = -4 $$:**
   $$
   x = 4
   $$
   
4. **Solution to $$ 4^{\log_2(x)} = 8 $$:**
   $$
   x = 2\sqrt{2}
   $$
**Domain of $$ f $$:** All real numbers. **Solutions:** - $$ f(x) \leq 0 $$ when $$ \frac{1}{4} \leq x \leq 1 $$. - $$ g(x) = -4 $$ when $$ x = 4 $$. - $$ 4^{\log_2(x)} = 8 $$ when $$ x = 2\sqrt{2} $$.

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