t (seconds) v (t) (feet/second) a (t) (feet/second 2) -20 -30 15 1 No. Since v (0) = V a (t) = v1 (t) = 0. O No. Since v (0) : - (+) (+) 5 25 -20 30 -14 1 35 -10 2 50 0 60 10 A car travels on a straight track. During that time interval 0 ≤ t ≤ 60 seconds, the car's velocity v, measured in feet per second, and acceleration a, measured in feet per second, are continuous functions. The table above shows the selected values of these functions. For 0 < t < 60, must there be a time a (t) = 0? Yes. Since v (0) = v (25), the Mean Value Theorem guarantees a t in (0,25) so that a (t) = v (t) = 0. 2 O Yes. Since v (0) = v (25), the Mean Value Theorem guarantees a t in (0,25) so that a (t) = v(t) = 0. (25), the Mean Value Theorem can not guarantee a t in (0,25) so that = v(25), the Mean Value Theorem can not guarantee a t in (0,25) so that

2. f (x) — — х? (х — 3) (х — 1) a. Find the x and y – intercepts : = - b. crosses or touches : c. End behavior : d. The maximum number or turning points : e. intervals : 100 50 f. -4 -2 2 4 -50

f(x) = x* + x³ – 2x? | For the above function state all intervals of x where f(x) is negative.

6e f(#)= 10 25e +10 then f(z) - Give the domain of f(r) in interval notation.