Got the samef'(t) = 4t(2t - 1) + (2t - 1)^2 (Using the Product Rule)
= 8t^2 - 4t + (2t - 1)^2
f"(t) = 16t - 4 + 4(2t - 1)
= 24t - 8
Therefore: f"(1) = 24(1) - 8
= 16
Your solution is wrong, f"(1) = 16
f''(1) = 196 was in the back of the textbook..f'(t) = 4t(2t - 1) + (2t - 1)^2 (Using the Product Rule)
= 8t^2 - 4t + (2t - 1)^2
f"(t) = 16t - 4 + 4(2t - 1)
= 24t - 8
Therefore: f"(1) = 24(1) - 8
= 16
Your solution is wrong, f"(1) = 16
Yeah, it is lol.Please don't tell me it is Maths In Focus.
If so, burn it.
Thank you. I see what I was doing wrong now...haha.I edited my original solution