consider the following.$ f(x) = 3x^4 - 12x^2 - 36 $$ \text{(a) find all real zeros of the polynomial function. (enter your answers as a comma-separated list. if there is no solution, enter no solution.)} $$ x = $$ \text{(b) determine the multiplicity of each zero.} $$ \text{smaller } x\text{-value} $$ \text{---select---} $$ \text{larger } x\text{-value} $$ \text{---select---} $$ \text{(c) determine the maximum possible number of turning points of the graph of the function.} $$ \text{turning point(s)} $$ \text{(d) use a graphing utility to graph the function and verify your answers.} $

Question

Accepted Answer

### Part (a)
# Explanation:
## Step1: Set $ f(x) = 0 $
We have the function $ f(x)=3x^{4}-12x^{2}-36 $. To find the real zeros, we set $ f(x) = 0 $, so:
$$
3x^{4}-12x^{2}-36 = 0
$$
## Step2: Divide by 3
Divide the entire equation by 3 to simplify:
$$
x^{4}-4x^{2}-12 = 0
$$
## Step3: Substitute $ u = x^{2} $
Let $ u=x^{2} $, then the equation becomes a quadratic in $ u $:
$$
u^{2}-4u - 12=0
$$
## Step4: Solve the quadratic equation
We factor the quadratic equation $ u^{2}-4u - 12 = 0 $. We need two numbers that multiply to - 12 and add to - 4. The numbers are - 6 and 2. So,
$$
(u - 6)(u+2)=0
$$
Setting each factor equal to zero gives $ u - 6=0 $ or $ u + 2=0 $, so $ u = 6 $ or $ u=-2 $.
## Step5: Substitute back $ u=x^{2} $
Since $ u=x^{2} $, we have two cases:
- Case 1: $ x^{2}=6 $, then $ x=\pm\sqrt{6}$
- Case 2: $ x^{2}=-2 $, since the square of a real number cannot be negative, there are no real solutions from this case.

# Answer:
$ -\sqrt{6},\sqrt{6} $

### Part (b)
For a zero $ x = a $ of a polynomial $ f(x)=(x - a)^{n}g(x) $ where $ g(a)
eq0 $, the multiplicity of $ a $ is $ n $.

- For $ x=-\sqrt{6} $: The factor corresponding to $ x = -\sqrt{6} $ comes from $ x^{2}-6=(x+\sqrt{6})(x - \sqrt{6}) $, so in the polynomial $ x^{4}-4x^{2}-12=(x^{2}-6)(x^{2}+2)=(x+\sqrt{6})(x - \sqrt{6})(x^{2}+2) $, the exponent of $ (x+\sqrt{6}) $ is 1. So the multiplicity of $ x = -\sqrt{6} $ is 1.
- For $ x=\sqrt{6} $: Similarly, the exponent of $ (x - \sqrt{6}) $ is 1. So the multiplicity of $ x=\sqrt{6} $ is 1.

# Brief Explanations:
The polynomial can be factored as $ (x + \sqrt{6})(x-\sqrt{6})(x^{2}+2) $. Each linear factor $ (x+\sqrt{6}) $ and $ (x - \sqrt{6}) $ has an exponent of 1, so the multiplicity of each zero $ -\sqrt{6} $ and $ \sqrt{6} $ is 1.
# Answer:
smaller $ x $-value ($ x = -\sqrt{6} $): multiplicity 1; larger $ x $-value ($ x=\sqrt{6} $): multiplicity 1

### Part (c)
# Explanation:
The degree of the polynomial $ f(x)=3x^{4}-12x^{2}-36 $ is $ n = 4 $. The maximum number of turning points of a polynomial function of degree $ n $ is $ n - 1 $.

## Step1: Recall the formula for turning points
For a polynomial function of degree $ n $, the maximum number of turning points is given by $ n-1 $.

## Step2: Substitute $ n = 4 $
Here, $ n = 4 $, so the maximum number of turning points is $ 4-1=3 $.

# Answer:
3

### Part (d)
To graph $ f(x)=3x^{4}-12x^{2}-36 $:

1. **End - behavior**: The leading term is $ 3x^{4} $. Since the degree $ 4 $ is even and the leading coefficient $ 3>0 $, as $ x
ightarrow\pm\infty $, $ f(x)
ightarrow\infty $.
2. **Zeros**: We know the real zeros are $ x = -\sqrt{6}\approx - 2.45 $ and $ x=\sqrt{6}\approx2.45 $.
3. **Turning points**: We can find the derivative $ f^\prime(x)=12x^{3}-24x = 12x(x^{2}-2)=12x(x + \sqrt{2})(x-\sqrt{2}) $. Setting $ f^\prime(x)=0 $ gives critical points at $ x = 0,x=-\sqrt{2}\approx - 1.41,x=\sqrt{2}\approx1.41 $. These are the $ x $-coordinates of the turning points. We can find the corresponding $ y $-values by plugging these $ x $-values into $ f(x) $:
   - $ f(0)=3(0)^{4}-12(0)^{2}-36=-36 $
   - $ f(-\sqrt{2})=3(-\sqrt{2})^{4}-12(-\sqrt{2})^{2}-36=3\times4-12\times2 - 36=12 - 24-36=-48 $
   - $ f(\sqrt{2})=3(\sqrt{2})^{4}-12(\sqrt{2})^{2}-36=3\times4-12\times2 - 36=12 - 24 - 36=-48 $

When we graph the function, we plot the zeros $ (-\sqrt{6},0)$ and $ (\sqrt{6},0) $, the turning points $ (0, - 36) $, $ (-\sqrt{2},-48) $, $ (\sqrt{2},-48) $ and use the end - behavior to sketch the curve. The graph will open upwards (since the leading coefficient of the even - degree polynomial is positive), have a "W" - like shape with the turning points at $ x = -\sqrt{2},0,\sqrt{2} $ and zeros at $ x=-\sqrt{6},\sqrt{6} $. This graph will verify our earlier results about zeros and turning points.

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step1: Set \( f(x) = 0 \)

Step2: Divide by 3

Step3: Substitute \( u = x^{2} \)

Step4: Solve the quadratic equation

Step5: Substitute back \( u=x^{2} \)

Step1: Recall the formula for turning points

Step2: Substitute \( n = 4 \)

Answer:

Part (b)