The graph of $y = x^3 - 7x - 12$ is shown below - OCR - GCSE Maths - Question 9 - 2018 - Paper 1

To show that $3 < p < 4$, we will evaluate $y$ at $x = 3$ and $x = 4$:

1. For $x = 3: $y = 3^3 - 7(3) - 12 = -6$

2. For $x = 4: $y = 4^3 - 7(4) - 12 = 64 - 28 - 12 = 24$

Since $y(3) = -6$ (a negative value) and $y(4) = 24$ (a positive value), there is a change of sign between $x = 3$ and $x = 4$. Therefore, by the Intermediate Value Theorem, it follows that:

$3 < p < 4$.

To find a smaller interval that contains $p$, we will evaluate $y$ at $x = 3.5$:

For $x = 3.5: $y = (3.5)^3 - 7(3.5) - 12$ Calculating this gives:

$= 42.875 - 24.5 - 12 = 6.375$

Now we have:

• $y(3) = -6$, which is negative.
• $y(3.5) = 6.375$, which is positive.

This also indicates a change of sign between $x = 3$ and $x = 3.5$, so we know:

$3 < p < 3.5$

Next, we check $x = 3.25$:

For $x = 3.25: $y = (3.25)^3 - 7(3.25) - 12$ Calculating:

$= 34.328125 - 22.75 - 12 = -0.421875$

Thus, we have:

• $y(3.25) = -0.421875$, still negative.
• $y(3.5) = 6.375$, positive.

Finally, we find: $3.25 < p < 3.5$

This is the smaller interval that contains $p$.