If fff is continuous on [0,3][0, 3][0,3] with f(0)=−2f(0) = -2f(0)=−2 and f(3)=5f(3) = 5f(3)=5, which value must fff attain by the IVT?
f(c)=1f(c) = 1f(c)=1 for some c∈(0,3)c \in (0,3)c∈(0,3)
f(c)=7f(c) = 7f(c)=7 for some c∈(0,3)c \in (0,3)c∈(0,3)
f(c)=−5f(c) = -5f(c)=−5 for some c∈(0,3)c \in (0,3)c∈(0,3)
f(c)=6f(c) = 6f(c)=6 for some c∈(0,3)c \in (0,3)c∈(0,3)