Question

name: aniwone dukresine
student number: 2537804
instructions:

1. no books or notes are permitted.
2. only calculators sharp el-531 are permitted.
3. please show all your work clearly and justify all your answers.
4. cheating will result in a minimum penalty of zero in your test.
5. please solve all limits algebraically, not numerically nor graphically.
6. 3 marks

find the extreme values of ( f(x) = x^6 - 6x^5 + 1 ) on the interval (1, 7).
( f(x) = x^6 - 6x^5 + 1 )
(1,7)
date: december 4, 2025
time: 10:05 - 11:35

Explanation:

Step1: Find the derivative of \( f(x) \)

To find the extreme values, we first find the derivative of \( f(x) = x^6 - 6x^5 + 1 \). Using the power rule, \( f'(x) = 6x^5 - 30x^4 \).

Step2: Find critical points

Set \( f'(x) = 0 \):

$$ 6x^5 - 30x^4 = 0 \\ 6x^4(x - 5) = 0 $$

This gives critical points at \( x = 0 \) and \( x = 5 \). But we are interested in the interval \([1, 7]\), so \( x = 5 \) is in the interval, and \( x = 0 \) is not.

Step3: Evaluate \( f(x) \) at critical points and endpoints

• Evaluate at \( x = 1 \):

$$ f(1) = 1^6 - 6(1)^5 + 1 = 1 - 6 + 1 = -4 $$

• Evaluate at \( x = 5 \):

$$ f(5) = 5^6 - 6(5)^5 + 1 = 15625 - 6(3125) + 1 = 15625 - 18750 + 1 = -3124 $$

• Evaluate at \( x = 7 \):

$$ f(7) = 7^6 - 6(7)^5 + 1 = 117649 - 6(16807) + 1 = 117649 - 100842 + 1 = 16808 $$

Step4: Determine extreme values

Compare the values: \( f(1) = -4 \), \( f(5) = -3124 \), \( f(7) = 16808 \). So the minimum value is \(-3124\) (at \( x = 5 \)) and the maximum value is \( 16808 \) (at \( x = 7 \)).

Answer:

The minimum value is \(\boldsymbol{-3124}\) (at \( x = 5 \)) and the maximum value is \(\boldsymbol{16808}\) (at \( x = 7 \)) on the interval \([1, 7]\).