a. Use the Intermediate Value Theorem to show that the following equation has a solution on the given interval.
b. Use a graphing utility to find all the solutions to the equation on the given interval.
c. Illustrate your answers with an appropriate graph.
\[
2 x^{3}+x-1=0 ;(0,1)
\]

Step 1 ：First, we need to understand the Intermediate Value Theorem (IVT). The IVT states that if a function \(f\) is continuous on a closed interval \([a, b]\), and \(k\) is any number between \(f(a)\) and \(f(b)\), then there exists a number \(c\) in the interval \([a, b]\) such that \(f(c) = k\).

Step 2 ：We are given the function \(f(x) = 2x^3 + x - 1\) and the interval \((0, 1)\). We need to show that there is a solution to the equation \(f(x) = 0\) in this interval.

Step 3 ：First, we calculate \(f(0)\) and \(f(1)\).

Step 4 ：Substitute \(x = 0\) into the equation, we get \(f(0) = 2(0)^3 + 0 - 1 = -1\).

Step 5 ：Substitute \(x = 1\) into the equation, we get \(f(1) = 2(1)^3 + 1 - 1 = 2\).

Step 6 ：Since \(f(0) = -1\) and \(f(1) = 2\), and 0 is between -1 and 2, by the Intermediate Value Theorem, there must be a value \(c\) in the interval \((0, 1)\) such that \(f(c) = 0\). So, the equation \(2x^3 + x - 1 = 0\) has a solution in the interval \((0, 1)\).

Step 7 ：Next, we use a graphing utility to find all the solutions to the equation on the given interval.

Step 8 ：By graphing the function \(f(x) = 2x^3 + x - 1\), we can see that the function crosses the x-axis at approximately \(x = 0.6527\) in the interval \((0, 1)\). So, the solution to the equation \(2x^3 + x - 1 = 0\) in the interval \((0, 1)\) is approximately \(x = 0.6527\).

Step 9 ：Finally, we illustrate our answers with an appropriate graph. The graph of the function \(f(x) = 2x^3 + x - 1\) shows that the function is continuous on the interval \((0, 1)\) and crosses the x-axis at approximately \(x = 0.6527\).