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Question 1
A drone's altitude (in metres) during a test flight is modelled by $h (t) = t^{3} - 6 t^{2} + 9 t + 2$ , where $t \geq 0$ is the time in seconds. During which time interval is the drone's altitude decreasing?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $1 < t < 3$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Differentiate $h(t)$
Find $h^{'} (t)$ to determine where the altitude is decreasing. $h^{'} (t) = 3 t^{2} - 12 t + 9$
Step 2: Solve $h'(t) = 0$
Set the derivative equal to zero: $3 t^{2} - 12 t + 9 = 0 ⟹ t^{2} - 4 t + 3 = 0 ⟹ (t - 1) (t - 3) = 0$ So the critical points are $t = 1$ and $t = 3$ .
Step 3: Test each interval
Test $t = 2$ in the interval $(1, 3)$ : $h^{'} (2) = 3 (4) - 12 (2) + 9 = 12 - 24 + 9 = - 3 < 0$ So $h^{'} (t) < 0$ on $(1, 3)$ , meaning the altitude is decreasing there.
Step 4: Confirm other intervals are increasing
Test $t = 0$ : $h^{'} (0) = 9 > 0$ (increasing). Test $t = 4$ : $h^{'} (4) = 48 - 48 + 9 = 9 > 0$ (increasing). The altitude is decreasing only on $1 < t < 3$ .
Method #2Approach 2
Step 1: Find the critical points
The critical points from $h^{'} (t) = 3 t^{2} - 12 t + 9 = 0$ are $t = 1$ and $t = 3$ . These bound any decreasing interval.
Step 2: Eliminate '$0 < t < 1$'
Testing $t = 0.5$ : $h^{'} (0.5) = 3 (0.25) - 6 + 9 = 3.75 > 0$ . The function is increasing here, so this option is wrong.
Step 3: Eliminate '$t > 3$'
Testing $t = 4$ : $h^{'} (4) = 48 - 48 + 9 = 9 > 0$ . The function is increasing for $t > 3$ , so this option is wrong.
Step 4: Eliminate '$0 < t < 3$'
This interval includes $(0, 1)$ where the function is increasing, so it cannot be the interval of decrease. This option is too broad and incorrect.
Step 5: Confirm '$1 < t < 3$'
Testing $t = 2$ : $h^{'} (2) = 12 - 24 + 9 = - 3 < 0$ . The function is decreasing on $(1, 3)$ , confirming this is the correct answer.
Question 2
The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 7$ models the temperature (°C) in a laboratory over time (hours). On which interval(s) is the temperature increasing?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $x < - 1$ and $x > 2$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Differentiate $f(x)$
$f^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x^{2} - x - 2) = 6 (x - 2) (x + 1)$
Step 2: Find critical points
Setting $f^{'} (x) = 0$ gives $x = 2$ or $x = - 1$ . These divide the number line into three intervals: $(- \infty, - 1)$ , $(- 1, 2)$ , and $(2, \infty)$ .
Step 3: Test each interval
Test $x = - 2$ : $f^{'} (- 2) = 6 (- 4) (- 1 + 1)$ ... more directly: $6 (- 2 - 2) (- 2 + 1) = 6 (- 4) (- 1) = 24 > 0$ ✓ increasing. Test $x = 0$ : $f^{'} (0) = 6 (0 - 2) (0 + 1) = - 12 < 0$ ✗ decreasing. Test $x = 3$ : $f^{'} (3) = 6 (1) (4) = 24 > 0$ ✓ increasing.
Step 4: State conclusion
The temperature is increasing on $x < - 1$ and $x > 2$ .
Method #2Approach 2
Step 1: Find critical points
$f^{'} (x) = 6 (x - 2) (x + 1) = 0$ gives $x = - 1$ and $x = 2$ . The sign of $f^{'}$ changes at these points.
Step 2: Eliminate '$-1 < x < 2$'
Test $x = 0$ : $f^{'} (0) = 6 (- 2) (1) = - 12 < 0$ . The function is decreasing on $(- 1, 2)$ , so this option is wrong.
Step 3: Eliminate '$x > 2$ only'
Test $x = - 2$ : $f^{'} (- 2) = 6 (- 4) (- 1) = 24 > 0$ . The function is also increasing for $x < - 1$ , so ' $x > 2$ only' is incomplete and wrong.
Step 4: Eliminate '$x < 2$ only'
This includes $(- 1, 2)$ where the function decreases. This option is clearly wrong.
Step 5: Confirm '$x < -1$ and $x > 2$'
Both intervals test positive for $f^{'} (x)$ , confirming the function increases on $x < - 1$ and $x > 2$ .

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Question 1
A drone's altitude (in metres) during a test flight is modelled by $h (t) = t^{3} - 6 t^{2} + 9 t + 2$ , where $t \geq 0$ is the time in seconds. During which time interval is the drone's altitude decreasing?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $1 < t < 3$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Differentiate $h(t)$
Find $h^{'} (t)$ to determine where the altitude is decreasing. $h^{'} (t) = 3 t^{2} - 12 t + 9$
Step 2: Solve $h'(t) = 0$
Set the derivative equal to zero: $3 t^{2} - 12 t + 9 = 0 ⟹ t^{2} - 4 t + 3 = 0 ⟹ (t - 1) (t - 3) = 0$ So the critical points are $t = 1$ and $t = 3$ .
Step 3: Test each interval
Test $t = 2$ in the interval $(1, 3)$ : $h^{'} (2) = 3 (4) - 12 (2) + 9 = 12 - 24 + 9 = - 3 < 0$ So $h^{'} (t) < 0$ on $(1, 3)$ , meaning the altitude is decreasing there.
Step 4: Confirm other intervals are increasing
Test $t = 0$ : $h^{'} (0) = 9 > 0$ (increasing). Test $t = 4$ : $h^{'} (4) = 48 - 48 + 9 = 9 > 0$ (increasing). The altitude is decreasing only on $1 < t < 3$ .
Method #2Approach 2
Step 1: Find the critical points
The critical points from $h^{'} (t) = 3 t^{2} - 12 t + 9 = 0$ are $t = 1$ and $t = 3$ . These bound any decreasing interval.
Step 2: Eliminate '$0 < t < 1$'
Testing $t = 0.5$ : $h^{'} (0.5) = 3 (0.25) - 6 + 9 = 3.75 > 0$ . The function is increasing here, so this option is wrong.
Step 3: Eliminate '$t > 3$'
Testing $t = 4$ : $h^{'} (4) = 48 - 48 + 9 = 9 > 0$ . The function is increasing for $t > 3$ , so this option is wrong.
Step 4: Eliminate '$0 < t < 3$'
This interval includes $(0, 1)$ where the function is increasing, so it cannot be the interval of decrease. This option is too broad and incorrect.
Step 5: Confirm '$1 < t < 3$'
Testing $t = 2$ : $h^{'} (2) = 12 - 24 + 9 = - 3 < 0$ . The function is decreasing on $(1, 3)$ , confirming this is the correct answer.
Question 2
The function $f (x) = 2 x^{3} - 3 x^{2} - 12 x + 7$ models the temperature (°C) in a laboratory over time (hours). On which interval(s) is the temperature increasing?
No clue? Show me the answer
Correct answerCorrect!Incorrect
A $x < - 1$ and $x > 2$
Step-by-step walkthrough
Choose a solution method
Method #1Approach 1
Step 1: Differentiate $f(x)$
$f^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x^{2} - x - 2) = 6 (x - 2) (x + 1)$
Step 2: Find critical points
Setting $f^{'} (x) = 0$ gives $x = 2$ or $x = - 1$ . These divide the number line into three intervals: $(- \infty, - 1)$ , $(- 1, 2)$ , and $(2, \infty)$ .
Step 3: Test each interval
Test $x = - 2$ : $f^{'} (- 2) = 6 (- 4) (- 1 + 1)$ ... more directly: $6 (- 2 - 2) (- 2 + 1) = 6 (- 4) (- 1) = 24 > 0$ ✓ increasing. Test $x = 0$ : $f^{'} (0) = 6 (0 - 2) (0 + 1) = - 12 < 0$ ✗ decreasing. Test $x = 3$ : $f^{'} (3) = 6 (1) (4) = 24 > 0$ ✓ increasing.
Step 4: State conclusion
The temperature is increasing on $x < - 1$ and $x > 2$ .
Method #2Approach 2
Step 1: Find critical points
$f^{'} (x) = 6 (x - 2) (x + 1) = 0$ gives $x = - 1$ and $x = 2$ . The sign of $f^{'}$ changes at these points.
Step 2: Eliminate '$-1 < x < 2$'
Test $x = 0$ : $f^{'} (0) = 6 (- 2) (1) = - 12 < 0$ . The function is decreasing on $(- 1, 2)$ , so this option is wrong.
Step 3: Eliminate '$x > 2$ only'
Test $x = - 2$ : $f^{'} (- 2) = 6 (- 4) (- 1) = 24 > 0$ . The function is also increasing for $x < - 1$ , so ' $x > 2$ only' is incomplete and wrong.
Step 4: Eliminate '$x < 2$ only'
This includes $(- 1, 2)$ where the function decreases. This option is clearly wrong.
Step 5: Confirm '$x < -1$ and $x > 2$'
Both intervals test positive for $f^{'} (x)$ , confirming the function increases on $x < - 1$ and $x > 2$ .

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Step-by-step walkthrough

Step 1: Differentiate $h(t)$

Step 2: Solve $h'(t) = 0$

Step 3: Test each interval

Step 4: Confirm other intervals are increasing

Step 1: Find the critical points

Step 2: Eliminate '$0 < t < 1$'

Step 3: Eliminate '$t > 3$'

Step 4: Eliminate '$0 < t < 3$'

Step 5: Confirm '$1 < t < 3$'

Step-by-step walkthrough

Step 1: Differentiate $f(x)$

Step 2: Find critical points

Step 3: Test each interval

Step 4: State conclusion

Step 1: Find critical points

Step 2: Eliminate '$-1 < x < 2$'

Step 3: Eliminate '$x > 2$ only'

Step 4: Eliminate '$x < 2$ only'

Step 5: Confirm '$x < -1$ and $x > 2$'

14 more questions in this topic

Step-by-step walkthrough

Step 1: Differentiate $h(t)$

Step 2: Solve $h'(t) = 0$

Step 3: Test each interval

Step 4: Confirm other intervals are increasing

Step 1: Find the critical points

Step 2: Eliminate '$0 < t < 1$'

Step 3: Eliminate '$t > 3$'

Step 4: Eliminate '$0 < t < 3$'

Step 5: Confirm '$1 < t < 3$'

Step-by-step walkthrough

Step 1: Differentiate $f(x)$

Step 2: Find critical points

Step 3: Test each interval

Step 4: State conclusion

Step 1: Find critical points

Step 2: Eliminate '$-1 < x < 2$'

Step 3: Eliminate '$x > 2$ only'

Step 4: Eliminate '$x < 2$ only'

Step 5: Confirm '$x < -1$ and $x > 2$'

14 more questions in this topic