Answer to Question #157998 in Calculus for Cypress

Question

Show that f(x)=(x^3)+(3x^2)-1 has at least one zero in each of the intervals [0,1],[-1,0] and [-3,-2]

Answer 1

For the proof, we will use the intermediate value theorem states that if f is a continuous function whose domain contains the interval [a, b], then it takes on any given value between f(a) and f(b) at some point within the interval.

This has two important corollaries:

1. If a continuous function has values of opposite sign inside an interval, then it has a root in that interval (Bolzano's theorem).
2. The image of a continuous function over an interval is itself an interval.

In our case,

1) $x\in[0,1]$

$$\left\{\begin{array}{l} f(0)=0^3+3\cdot 0^2-1=-1<0\\[0.3cm] f(1)=1^3+3\cdot 1^2-1=3>0 \end{array}\right.\longrightarrow\\[0.3cm] \exists c\in(0,1) : f(c)=0$$

2) $x\in[-1,0]$

$$\left\{\begin{array}{l} f(0)=0^3+3\cdot 0^2-1=-1<0\\[0.3cm] f(-1)=(-1)^3+3\cdot(-1)^2-1=1>0 \end{array}\right.\longrightarrow\\[0.3cm] \exists c\in(-1,0) : f(c)=0$$

3) $x\in[-3,-2]$

$$\left\{\begin{array}{l} f(-3)=(-3)^3+3\cdot(-3)^2-1=-1<0\\[0.3cm] f(-2)=(-2)^3+3\cdot(-2)^2-1=3>0 \end{array}\right.\longrightarrow\\[0.3cm] \exists c\in(-3,-2) : f(c)=0$$

Q.E.D.