I assume (x-4)3 means (x-4)³. What we wish is to set the derivative equal to zero.
Expanding the T(x) polynomial makes it easier for me to take the derivative.
So (x-4)³ = x³ - 12x² + 48x - 64 + 6
T'(x) = 3x² - 24x + 48
We can factor out a 3 and set this to zero:
x² - 8x + 16 = 0
(x -4)² = 0
x = 4 should therefore represent the turning point.
I am mildly chagrined, I almost used the f'(x) = nx^(n-1) function at first, which appears would have been correct.
Answer:
B
Step-by-step explanation:
Have a nice day!
Answer:
the answer is the 2nd one
Step-by-step explanation:
Answer:
5
Step-by-step explanation:
Subtract 1 from both sides. The equation becomes 2.5d + 6.25 = 3.75d
Subtract 2.5d from both sides. The equation becomes 6.25 = 1.25d
Divide 1.25 from both sides. You get 5 = d
