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Find the $x$-coordinates of the stationary points on the curve with equation y = \frac{1}{2}x^3 - 2x^2 + 6. - Scottish Highers Maths - Question 1 - 2022

Question 1

$Find-the-$x$-coordinates-of-the-stationary-points-on-the-curve-with-equation--y-=-\frac{1}{2}x^3---2x^2-+-6.-Scottish Highers Maths-Question 1-2022.png$

Find the $x$-coordinates of the stationary points on the curve with equation y = \frac{1}{2}x^3 - 2x^2 + 6.

Worked Solution & Example Answer:Find the $x$-coordinates of the stationary points on the curve with equation y = \frac{1}{2}x^3 - 2x^2 + 6. - Scottish Highers Maths - Question 1 - 2022

Step 1

Step 1: Differentiate the function

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Answer

To find the stationary points, we first need to differentiate the function. The derivative of $y$ with respect to $x$ is obtained using the power rule:

$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}x^3 - 2x^2 + 6\right) = \frac{3}{2}x^2 - 4x.$

Step 2

Step 2: Set the derivative to zero

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Answer

To find stationary points, set the derivative equal to zero:

$\frac{3}{2}x^2 - 4x = 0.$

Factoring out the common term gives:

$x(\frac{3}{2}x - 4) = 0.$

Step 3

Step 3: Solve for $x$

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Answer

Now we can solve for $x$ :

Setting $x = 0$ .
Setting $\frac{3}{2}x - 4 = 0$ leads to: $\frac{3}{2}x = 4\Rightarrow x = \frac{4 \cdot 2}{3} = \frac{8}{3}.$

Thus, the $x$ -coordinates of the stationary points are:

$x = 0 \quad \text{and} \quad x = \frac{8}{3}.$

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