Answer
Apply the Intermediate Value Theorem and show that the line $y=0$ must pass the line $y=f(x)$ at least once on the interval $[0,1]$.
Work Step by Step
According to the Intermediate Value Theorem: if $f$ is a continuous function on a closed interval $[a,b]$, and if $y_0$ is any value between $f(a)$ and $f(b)$, then there exists some $c$ in $[a,b]$ such that $f(c)=y_0$
Apply the Theorem to this exercise, we have:
- $y=f(x)$ is a continuous function.
- At $x=0$, $f(0)\lt0$
- At $x=1$, $f(1)\gt0$
Therefore, because $y_0=0$ is a value between $f(0)$ and $f(1)$, there must exist at least a value of $x=c\in[0,1]$ such that $f(c)=0$. In other words, $f(x)=0$ has at least one solution between $x=0$ and $x=1$.
We can conceive this exercise as well as this theorem graphically.
- Because $y=f(x)$ is a continuous function on $[0,1]$, its graph is a no-break continuous line going from $f(0)$ to $f(1)$. And since $f(0)$ is negative and $f(1)$ is positive, this line must pass the x-axis at some point while going from $x=0$ to $x=1$.
- Or we might say the line $y=0$ must intersect the line $y=f(x)$ sooner or later as $x\in[0,1]$
The graph below might help to conceive this problem better. The graph portrays the line $f(x)=x-1/2$. You can see that as $x$ goes from $0$ to $1$, $f(x)$ passes the x-axis at $x=1/2$.
