Term 241, MATH 101, Makeup Quiz 2 (1) Page 2 of 5 3. By using the intermediate value theorem, the two curves \( f(x)=x^{4}-x^{2}+4 x \) and \( g(x)=x+1 \) intersect in the interval

To find the intersection points of the curves \( f(x) = x^4 - x^2 + 4x \) and \( g(x) = x + 1 \), we need to set them equal to each other: \[ f(x) = g(x) \] This gives us the equation: \[ x^4 - x^2 + 4x = x + 1 \] Rearranging this equation, we get: \[ x^4 - x^2 + 4x - x - 1 = 0 \] Simplifying further: \[ x^4 - x^2 + 3x - 1 = 0 \] Now, we will use the Intermediate Value Theorem (IVT) to find an interval where a root exists. The IVT states that if a continuous function changes signs over an interval \([a, b]\), then there is at least one root in that interval. Let's define a new function: \[ h(x) = x^4 - x^2 + 3x - 1 \] Next, we will evaluate \( h(x) \) at several points to find an interval where it changes sign. 1. Evaluate \( h(0) \): \[ h(0) = 0^4 - 0^2 + 3(0) - 1 = -1 \] 2. Evaluate \( h(1) \): \[ h(1) = 1^4 - 1^2 + 3(1) - 1 = 1 - 1 + 3 - 1 = 2 \] 3. Evaluate \( h(-1) \): \[ h(-1) = (-1)^4 - (-1)^2 + 3(-1) - 1 = 1 - 1 - 3 - 1 = -4 \] 4. Evaluate \( h(2) \): \[ h(2) = 2^4 - 2^2 + 3(2) - 1 = 16 - 4 + 6 - 1 = 17 \] Now we have the following evaluations: - \( h(0) = -1 \) (negative) - \( h(1) = 2 \) (positive) - \( h(-1) = -4 \) (negative) - \( h(2) = 17 \) (positive) From these evaluations, we can see that: - \( h(0) < 0 \) and \( h(1) > 0 \) indicates that there is at least one root in the interval \((0, 1)\). - \( h(-1) < 0 \) and \( h(1) > 0 \) indicates that there is at least one root in the interval \((-1, 1)\). Thus, the two curves \( f(x) \) and \( g(x) \) intersect in the intervals \((-1, 1)\) and \((0, 1)\). To summarize, the curves intersect in the interval: \[ (-1, 1) \]