Finding Values for Which $$ f(x)=0 $$ In Exercises $$ 35, \underline{36}, \underline{37}, \underline{38}, \underline{39}, \underline{40}, \underline{41} $$, and $$ \underline{42} $$, fi values of $$ x $$ for which $$ f(x)=0 $$. 35. $$ f(x)=15-3 x $$ Answer $$ \downarrow $$ 36. $$ f(x)=4 x+6 $$ 37. $$ f(x)=\frac{3 x-4}{5} $$ Answer $$ \boldsymbol{*} $$ 38. $$ f(x)=\frac{12-x^{2}}{8} $$ 39. $$ f(x)=x^{2}-81 $$ Answer * 40. $$ f(x)=z^{2}-6 x-16 $$ 41. $$ f(x)=x^{3}-x $$

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Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{\left(3x-4\right)}{5}=0$$
- step1: Remove the parentheses:
  $$\frac{3x-4}{5}=0$$
- step2: Simplify:
  $$3x-4=0$$
- step3: Move the constant to the right side:
  $$3x=0+4$$
- step4: Remove 0:
  $$3x=4$$
- step5: Divide both sides:
  $$\frac{3x}{3}=\frac{4}{3}$$
- step6: Divide the numbers:
  $$x=\frac{4}{3}$$
Solve the equation $$ x^{2}-81=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve using square roots:
  $$x^{2}-81=0$$
- step1: Add or subtract both sides:
  $$x^{2}=81$$
- step2: Simplify the expression:
  $$x=\pm \sqrt{81}$$
- step3: Simplify the expression:
  $$x=\pm 9$$
- step4: Separate into possible cases:
  $$\begin{align}&x=9\&x=-9\end{align}$$
- step5: Rewrite:
  $$x_{1}=-9,x_{2}=9$$
Solve the equation $$ x^{3}-x=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$x^{3}-x=0$$
- step1: Factor the expression:
  $$x\left(x^{2}-1\right)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&x=0\&x^{2}-1=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=0\&x=1\&x=-1\end{align}$$
- step4: Rewrite:
  $$x_{1}=-1,x_{2}=0,x_{3}=1$$
Solve the equation $$ 15-3x=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$15-3x=0$$
- step1: Move the constant to the right side:
  $$-3x=0-15$$
- step2: Remove 0:
  $$-3x=-15$$
- step3: Change the signs:
  $$3x=15$$
- step4: Divide both sides:
  $$\frac{3x}{3}=\frac{15}{3}$$
- step5: Divide the numbers:
  $$x=5$$
Solve the equation $$ 4x+6=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$4x+6=0$$
- step1: Move the constant to the right side:
  $$4x=0-6$$
- step2: Remove 0:
  $$4x=-6$$
- step3: Divide both sides:
  $$\frac{4x}{4}=\frac{-6}{4}$$
- step4: Divide the numbers:
  $$x=-\frac{3}{2}$$
Solve the equation $$ z^{2}-6x-16=0 $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$z^{2}-6x-16=0$$
- step1: Rewrite the expression:
  $$z^{2}-16-6x=0$$
- step2: Move the expression to the right side:
  $$-6x=0-\left(z^{2}-16\right)$$
- step3: Subtract the terms:
  $$-6x=-z^{2}+16$$
- step4: Change the signs:
  $$6x=z^{2}-16$$
- step5: Divide both sides:
  $$\frac{6x}{6}=\frac{z^{2}-16}{6}$$
- step6: Divide the numbers:
  $$x=\frac{z^{2}-16}{6}$$
Solve the equation $$ (12-x^{2})/8=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve using square roots:
  $$\frac{\left(12-x^{2}\right)}{8}=0$$
- step1: Remove the parentheses:
  $$\frac{12-x^{2}}{8}=0$$
- step2: Multiply both sides:
  $$12-x^{2}=0$$
- step3: Add or subtract both sides:
  $$-x^{2}=-12$$
- step4: Divide both sides:
  $$\frac{-x^{2}}{-1}=\frac{-12}{-1}$$
- step5: Divide the numbers:
  $$x^{2}=12$$
- step6: Simplify the expression:
  $$x=\pm \sqrt{12}$$
- step7: Simplify the expression:
  $$x=\pm 2\sqrt{3}$$
- step8: Separate into possible cases:
  $$\begin{align}&x=2\sqrt{3}\&x=-2\sqrt{3}\end{align}$$
- step9: Rewrite:
  $$x_{1}=-2\sqrt{3},x_{2}=2\sqrt{3}$$
Here are the values of $$ x $$ for which $$ f(x) = 0 $$ for each of the given functions:

35. For $$ f(x) = 15 - 3x $$:
   $$
   15 - 3x = 0 \implies 3x = 15 \implies x = 5
   $$
   **Answer:** $$ x = 5 $$

36. For $$ f(x) = 4x + 6 $$:
   $$
   4x + 6 = 0 \implies 4x = -6 \implies x = -\frac{3}{2} \quad \text{or} \quad x = -1.5
   $$
   **Answer:** $$ x = -\frac{3}{2} $$

37. For $$ f(x) = \frac{3x - 4}{5} $$:
   $$
   \frac{3x - 4}{5} = 0 \implies 3x - 4 = 0 \implies 3x = 4 \implies x = \frac{4}{3} \quad \text{or} \quad x = 1.\dot{3}
   $$
   **Answer:** $$ x = \frac{4}{3} $$

38. For $$ f(x) = \frac{12 - x^2}{8} $$:
   $$
   \frac{12 - x^2}{8} = 0 \implies 12 - x^2 = 0 \implies x^2 = 12 \implies x = -2\sqrt{3} \quad \text{or} \quad x = 2\sqrt{3}
   $$
   **Answer:** $$ x_{1} = -2\sqrt{3}, x_{2} = 2\sqrt{3} $$

39. For $$ f(x) = x^2 - 81 $$:
   $$
   x^2 - 81 = 0 \implies x^2 = 81 \implies x = -9 \quad \text{or} \quad x = 9
   $$
   **Answer:** $$ x_{1} = -9, x_{2} = 9 $$

40. For $$ f(x) = z^2 - 6x - 16 $$:
   $$
   z^2 - 6x - 16 = 0 \implies 6x = z^2 - 16 \implies x = \frac{z^2 - 16}{6}
   $$
   **Answer:** $$ x = \frac{z^2 - 16}{6} $$

41. For $$ f(x) = x^3 - x $$:
   $$
   x^3 - x = 0 \implies x(x^2 - 1) = 0 \implies x(x - 1)(x + 1) = 0
   $$
   Thus, $$ x = -1, 0, 1 $$.
   **Answer:** $$ x_{1} = -1, x_{2} = 0, x_{3} = 1 $$

These are the values of $$ x $$ for which $$ f(x) = 0 $$ for each exercise.
Here are the values of $$ x $$ for which $$ f(x) = 0 $$ for each function: 35. $$ x = 5 $$ 36. $$ x = -1.5 $$ 37. $$ x = 1.\dot{3} $$ 38. $$ x = -2\sqrt{3} $$ or $$ x = 2\sqrt{3} $$ 39. $$ x = -9 $$ or $$ x = 9 $$ 40. $$ x = \frac{z^2 - 16}{6} $$ 41. $$ x = -1 $$, $$ x = 0 $$, or $$ x = 1 $$ These are the solutions where each function equals zero.

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