The diagram below shows two functions $f(x)$ and $g(x)$ - Leaving Cert Mathematics - Question 4 - 2020

Firstly, we need to find the turning points of the function $f(x)$. This occurs when $f'(x) = 0$:

$f'(x) = 3x^2 + 2(-9)x + 15 = 0$

This simplifies to:

$3x^2 - 18x + 15 = 0$ $x^2 - 6x + 5 = 0$

Factoring the quadratic equation:

$(x - 5)(x - 1) = 0$

The turning points are at $x = 5$ and $x = 1$. Now, finding the corresponding y-values:

$f(5) = 5^3 - 9(5^2) + 15(5) + 8 = 15$ $f(1) = 1^3 - 9(1^2) + 15(1) + 8 = -17$

Thus, the turning points are $(1, -17)$ and $(5, 15)$. Now we calculate the slope of the line $g(x)$ that passes through these points:

$m_{g(x)} = \frac{15 - (-17)}{5 - 1} = \frac{32}{4} = 8$

Using the point-slope form for the line $g(x)$, taking point $(1, -17)$:

$y - (-17) = 8(x - 1)$

Expanding this gives:

$y + 17 = 8x - 8$ $y = 8x - 25$

Thus, the equation of $g(x)$ is:

$g(x) = 8x - 25$

The point of inflection occurs where the second derivative changes sign, which is defined by:

$f''(x) = 6x - 18$

Setting $f''(x) = 0$ to find the x-coordinate:

$6x - 18 = 0\Rightarrow x = 3$

Now, we need to find the corresponding y-value by substituting $x = 3$ back into $f(x)$:

$f(3) = 3^3 - 9(3^2) + 15(3) + 8 = -1$

Thus, the point of inflection is $(3, -1)$. Now, substituting $x = 3$ into $g(x)$ to check if this point is on the graph:

$g(3) = 8(3) - 25 = 24 - 25 = -1$

Since $g(3) = -1$, we confirm:

• The point of inflection $(3, -1)$ indeed lies on the graph of $g(x)$.