4 Show that the maximum value of $$ \sqrt{-x^{2}+4 x+12} $$ is 4.

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1. Begin with the expression:
   $$
   \sqrt{-x^{2}+4x+12}
   $$

2. Identify the quadratic inside the square root:
   $$
   f(x) = -x^{2}+4x+12
   $$

3. Factor out $$-1$$ from the terms involving $$x$$:
   $$
   f(x) = -\left(x^{2}-4x\right)+12
   $$

4. Complete the square for the quadratic $$x^{2}-4x$$:
   - Take half of the coefficient of $$x$$: $$-\frac{4}{2} = -2$$.
   - Square it: $$(-2)^2 = 4$$.
   - Add and subtract $$4$$ inside the parenthesis:
     $$
     x^{2}-4x = \left(x^{2}-4x+4\right)-4 = (x-2)^2-4
     $$

5. Substitute back into $$f(x)$$:
   $$
   f(x) = -\left[(x-2)^2-4\right]+12
   $$
   Simplify:
   $$
   f(x) = -(x-2)^2+4+12 = -(x-2)^2+16
   $$

6. The expression for $$f(x)$$ is now:
   $$
   f(x) = 16 - (x-2)^2
   $$
   Since $$(x-2)^2$$ is always non-negative, the maximum value of $$f(x)$$ occurs when:
   $$
   (x-2)^2 = 0 \quad \text{or} \quad x=2
   $$
   Thus:
   $$
   f(2) = 16 - 0 = 16
   $$

7. Since we are dealing with the square root of $$f(x)$$, take the square root of the maximum value:
   $$
   \sqrt{f(2)} = \sqrt{16} = 4
   $$

Therefore, the maximum value of $$\sqrt{-x^{2}+4x+12}$$ is $$4$$.
The maximum value of $$ \sqrt{-x^{2}+4x+12} $$ is 4.

4 Show that the maximum value of is 4.

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