20 12 The equation of a curve is $$ y=x^{4}-8 x^{2}+5 $$. (a) Find the derivative, $$ \left(\frac{d y}{d x}\right) $$, of $$ y=x^{4}-8 x^{2}+5 $$. [2] (b) Find the coordinates of the three turning points. You must show all your working. $$ \qquad $$ and ( $$ \qquad $$ ) and ( $$ \qquad $$ [4] (c) Determine which one of these turning points is a maximum. Justify your answer.

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### Part (a): Find the derivative $$ \frac{dy}{dx} $$

The given equation is:

$$
y = x^4 - 8x^2 + 5
$$

To find the derivative, we will differentiate each term with respect to $$ x $$:

1. The derivative of $$ x^4 $$ is $$ 4x^3 $$.
2. The derivative of $$ -8x^2 $$ is $$ -16x $$.
3. The derivative of the constant $$ 5 $$ is $$ 0 $$.

Putting it all together, we have:

$$
\frac{dy}{dx} = 4x^3 - 16x
$$

### Part (b): Find the coordinates of the three turning points

Turning points occur where the derivative is equal to zero. We set the derivative equal to zero:

$$
4x^3 - 16x = 0
$$

Factoring out the common term:

$$
4x(x^2 - 4) = 0
$$

This gives us:

$$
4x = 0 \quad \text{or} \quad x^2 - 4 = 0
$$

From $$ 4x = 0 $$, we get:

$$
x = 0
$$

From $$ x^2 - 4 = 0 $$, we get:

$$
x^2 = 4 \implies x = 2 \quad \text{or} \quad x = -2
$$

Thus, the $$ x $$-coordinates of the turning points are $$ x = -2, 0, 2 $$.

Next, we find the corresponding $$ y $$-coordinates by substituting these $$ x $$-values back into the original equation:

1. For $$ x = -2 $$:

$$
y = (-2)^4 - 8(-2)^2 + 5 = 16 - 32 + 5 = -11
$$

2. For $$ x = 0 $$:

$$
y = 0^4 - 8(0)^2 + 5 = 5
$$

3. For $$ x = 2 $$:

$$
y = (2)^4 - 8(2)^2 + 5 = 16 - 32 + 5 = -11
$$

Thus, the coordinates of the turning points are:

- $$ (-2, -11) $$
- $$ (0, 5) $$
- $$ (2, -11) $$

### Part (c): Determine which one of these turning points is a maximum

To determine whether each turning point is a maximum or minimum, we can use the second derivative test. We first find the second derivative $$ \frac{d^2y}{dx^2} $$.

Starting from the first derivative:

$$
\frac{dy}{dx} = 4x^3 - 16x
$$

Now, we differentiate again:

1. The derivative of $$ 4x^3 $$ is $$ 12x^2 $$.
2. The derivative of $$ -16x $$ is $$ -16 $$.

Thus, the second derivative is:

$$
\frac{d^2y}{dx^2} = 12x^2 - 16
$$

Now we evaluate the second derivative at each turning point:

1. For $$ x = -2 $$:

$$
\frac{d^2y}{dx^2} = 12(-2)^2 - 16 = 12(4) - 16 = 48 - 16 = 32 \quad (\text{positive, so minimum})
$$

2. For $$ x = 0 $$:

$$
\frac{d^2y}{dx^2} = 12(0)^2 - 16 = 0 - 16 = -16 \quad (\text{negative, so maximum})
$$

3. For $$ x = 2 $$:

$$
\frac{d^2y}{dx^2} = 12(2)^2 - 16 = 12(4) - 16 = 48 - 16 = 32 \quad (\text{positive, so minimum})
$$

### Conclusion

The turning point that is a maximum is at:

$$
(0, 5)
$$

This is justified because the second derivative at this point is negative, indicating a local maximum.
The turning point at $$ (0, 5) $$ is a maximum.

20

12 The equation of a curve is .
(a) Find the derivative, , of .
[2]
(b) Find the coordinates of the three turning points.

You must show all your working.
and ( ) and ( [4]
© Determine which one of these turning points is a maximum.

Justify your answer.

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20 12 The equation of a curve is . (a) Find the derivative, , of . [2] (b) Find the coordinates of the three turning points. You must show all your working. and ( ) and ( [4] © Determine which one of these turning points is a maximum. Justify your answer.