Quiz 2 - Continuity

Problem 1

Question

Show that the graph of \(f(x) = 3x^2 + 5x - 11 = 0\) has a solution between \(x=1\) and \(x=2\). State which theorem you used.

Solution

Since we know that \(f(x)\) is a polynomial and thus continous, we can use the intermediate value theorem to show that there exists a root between the points. To do so we need to show that the function is above the x-axis at one of the endpoints and below at the other: \[ \begin{align*} f(1) &= 3(1)^2 + 5(1) - 11 = 3 + 5 - 11 = -3 < 0 \\ f(2) &= 3(2)^2 + 5(2) - 11 = 12 + 10 - 11 = 11 > 0 \end{align*} \]

Problem 2

Question

Find constants \(a,b\) so that the function given below is continuous for all \(x\): \[ f(x) = \begin{cases} x^2 + 3 & x<2 \\ a & x = 2 \\ ax+b & x > 2 \end{cases} \]

Solution

First, we need to make sure that the left most part of the function connects with the center part: \[ \begin{align*} a &= x^2 + 3 \text{ at }x=2 \\ &= (2)^2 + 3 \\ \Aboxed{a &= 7} \end{align*} \] Next, we match the center part to the rightmost part: \[ \begin{align*} a &= ax + b\text{ at }x=2 \\ 7 &= 7(2) + b \\ b &= 7 - 14 \\ \Aboxed{b &= -7} \end{align*} \]