Answer:
True, see proof below.
Step-by-step explanation:
Remember two theorems about continuity:
- If f is differentiable at the point p, then f is continuous at p. This also applies to intervals instead of points.
- (Bolzano) If f is continuous in an interval [a,b] and there exists x,y∈[a,b] such that f(x)<0<f(y), then there exists some c∈[a,b] such that f(c)=0.
If f is differentiable in [0,4], then f is continuous in [0,4] (by 1). Now, f(0)=-1<0 and f(4)=3>0. Thus, we have the inequality f(0)<0<f(4). By Bolzano's theorem, there exists some c∈[0,4] such that f(c)=0.
The answer to your question is (A, -3,-5)
The best way to answer this is to find counterarguments for each answer to eliminate them as a possibility. ab doesn't have to be an integer, 2.5*3=7.5 . That's also not an imaginary number so that's out. And that's also not an irrational number so that's out too. Only choice left is rational number.
