Practice problem
Given the function k(t)=t3+2k(t)=t^3+2k(t)=t3+2:
a) Evaluate k(2)k(2)k(2).
b) Solve k(t)=1k(t) = 1k(t)=1.
a) k(2)=8k(2)=8k(2)=8
a) k(2)=10k(2)=10k(2)=10
a) k(2)=6k(2)=6k(2)=6
a) k(2)=4k(2)=4k(2)=4
b) t=−1t=-1t=−1
b) t=1t=1t=1
b) t=−33t=-\sqrt[3]{3}t=−33
b) t=−2t=-2t=−2
Type a concept, like “slope” or “photosynthesis”.