Question

1. use the intermediate value theorem to find the value of c in the interval. $f(x)=2x^{2}+3x + 5$ $1,4$, $f(c)=19$

Explanation:

Step1: Evaluate function at endpoints

First, find \(f(1)\) and \(f(4)\).
For \(x = 1\), \(f(1)=2\times1^{2}+3\times1 + 5=2 + 3+5=10\).
For \(x = 4\), \(f(4)=2\times4^{2}+3\times4 + 5=2\times16+12 + 5=32+12 + 5=49\).
Since \(y = 19\) and \(10<19<49\), by the intermediate - value theorem, there exists a \(c\in[1,4]\) such that \(f(c)=19\).

Step2: Set up the equation

Set \(f(c)=2c^{2}+3c + 5 = 19\).
Rearrange to get a quadratic equation: \(2c^{2}+3c+5 - 19=0\), so \(2c^{2}+3c - 14 = 0\).

Step3: Solve the quadratic equation

For a quadratic equation \(ax^{2}+bx + c = 0\) (here \(a = 2\), \(b = 3\), \(c=-14\)), use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).
\(c=\frac{-3\pm\sqrt{3^{2}-4\times2\times(-14)}}{2\times2}=\frac{-3\pm\sqrt{9 + 112}}{4}=\frac{-3\pm\sqrt{121}}{4}=\frac{-3\pm11}{4}\).
We get two solutions: \(c_1=\frac{-3 + 11}{4}=\frac{8}{4}=2\) and \(c_2=\frac{-3 - 11}{4}=\frac{-14}{4}=-\frac{7}{2}\).

Step4: Select the valid solution

Since \(c\in[1,4]\), we reject \(c =-\frac{7}{2}\).

Answer:

\(c = 2\)