# Video: APCALC02AB-P1A-Q09-312192865392

A function 𝑓 is continuous on [−2, 2] and some values of 𝑓 are shown in the table. If 𝑓(𝑥) = 0 has only one solution, 𝑟, and 𝑟 < 0, what is a possible value of 𝑘? [A] 1 [B] 0 [C] 2 [D] −1

02:12

### Video Transcript

A function 𝑓 is continuous on the closed interval negative two to two and some values of 𝑓 are shown in the table. If 𝑓 of 𝑥 equals zero has only one solution, 𝑟, and 𝑟 is less than zero, what is a possible value of 𝑘? Is it a) one, b) zero, c) two, or d) negative one.

Let’s think about all the detail in this question and what it’s really telling us. We know that 𝑓 of 𝑥 equals zero has one solution, 𝑟. So this tells us that 𝑓 of 𝑟 equals zero. We also know that 𝑟 itself is less than zero. And the function is continuous over the closed interval negative two to two. We can also use the values given in our table to say that 𝑓 of negative two is equal to three. 𝑓 of zero is equal to 𝑘. And 𝑓 of positive two is equal to negative three. So where does 𝑓 of 𝑟 fit in here?

We know that 𝑓 of 𝑟 is equal to zero. So let’s sketch what our function could look like. 𝑓 of negative two is equal to three. That’s represented by the point whose Cartesian coordinates are negative two, three. We know that 𝑓 of two is equal to negative three. So we can say that that has Cartesian coordinates two, negative three. 𝑓 of zero is 𝑘. But we can’t add this just yet. We can however add 𝑓 of 𝑟. 𝑓 of 𝑟 is zero. And 𝑟 itself is less than zero. The point that represents this is going to be on the negative side of our 𝑥-axis.

And since 𝑓 of 𝑥 equals zero has only one solution, 𝑟, we know that this is the only point at which our function crosses the 𝑥-axis. So our function could look as shown. 𝑓 of zero is the point at which our curve crosses the 𝑦-axis. It has Cartesian coordinates zero, 𝑓 of zero or zero, 𝑘. Since this point sits on the negative half of the 𝑦-axis, we know that 𝑘 must be less than zero.

The only option in this list that satisfies this criteria is d. If 𝑓 of 𝑥 equals zero has one solution, 𝑟, and 𝑟 is less than zero, the only possible value of 𝑘 here is negative one.