Hannah is doing a training session - Leaving Cert Mathematics - Question 7 - 2022

To find h′(x), we differentiate h(x):

$h(x) = 2x^3 - 28x^2 + 105x + 70$

Using power rule:

$h′(x) = 6x^2 - 56x + 105$

Substituting x = 2:

$h′(2) = 6(2)^2 - 56(2) + 105$

Calculating:

1. $6(2)^2 = 24$.
2. $-56(2) = -112$.
3. Adding up: $24 - 112 + 105 = 17$.

Therefore, $h′(2) = 17$. This means that at 2 minutes, Hannah’s heart-rate is increasing at a rate of 17 BPM per minute.

To find the least and greatest values, evaluate the critical points from $h′(x) = 0$:

$6x^2 - 56x + 105 = 0$

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{56 \pm \sqrt{(-56)^2 - 4(6)(105)}}{2(6)}$

This gives: $x = \frac{56 \pm \sqrt{3136 - 2520}}{12} = \frac{56 \pm \sqrt{616}}{12} \\x = \frac{56 \pm 24.8}{12}$

Finding solutions, evaluate h(x) at critical points and the endpoints:

1. Calculate h(0), h(8), and values at critical points to find the least and greatest.

To find where Hannah’s heart-rate decreases most quickly, evaluate the second derivative:

$h′′(x) = 12x - 56$

Setting $h′′(x) = 0$:

$12x - 56 = 0$

Solving gives: $x = \frac{56}{12} = 4.67$

To find the time in minutes and seconds, convert:

0.67 minutes = 40 seconds.

Thus, heart-rate decreases most quickly after approximately 4 minutes and 40 seconds.

Bruno's heart-rate can be expressed as:

$b(x) = h(x) + 15$

Thus: $b(x) = (2x^3 - 28x^2 + 105x + 70) + 15 = 2x^3 - 28x^2 + 105x + 85$

Karen's heart-rate can be expressed as:

$k(x) = 0.9h(x)$

This gives: $k(x) = 0.9(2x^3 - 28x^2 + 105x + 70) = 1.8x^3 - 25.2x^2 + 94.5x + 63$

To express Martha's heart-rate:

$m(x) = h(0.8x)$

Substituting gives:

$m(x) = 2(0.8x)^3 - 28(0.8x)^2 + 105(0.8x) + 70$

After calculations: $m(x) = 2(0.512x^3) - 28(0.64x^2) + 84x + 70$

Thus, it simplifies to:

$m(x) = 1.024x^3 - 17.92x^2 + 84x + 70$