7 Seja $f(x)=3 x^{2}-2 x+m$.
a) Determine o valor de $m$ para que o mínimo seja $\frac{4}{3}$.
b) Para o valor de $m$ obtido no item anterior, determine $x$ para que $f(x)=\frac{4}{3}$

Step 1 ：\(f(x) = 3x^2 - 2x + m\)

Step 2 ：To find the minimum value of \(f(x)\), we can find the vertex of the parabola. The x-coordinate of the vertex is given by \(x_v = \frac{-b}{2a}\), where \(a = 3\) and \(b = -2\).

Step 3 ：\(x_v = \frac{-(-2)}{2(3)} = \frac{2}{6} = \frac{1}{3}\)

Step 4 ：Now, we can find the y-coordinate of the vertex by plugging \(x_v\) into the function \(f(x)\):

Step 5 ：\(y_v = f\left(\frac{1}{3}\right) = 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right) + m\)

Step 6 ：\(y_v = 3\left(\frac{1}{9}\right) - \frac{2}{3} + m = \frac{1}{3} - \frac{2}{3} + m = \frac{4}{3}\)

Step 7 ：Now, we can solve for \(m\):

Step 8 ：\(\frac{4}{3} = \frac{1}{3} - \frac{2}{3} + m\)

Step 9 ：\(\frac{4}{3} = -\frac{1}{3} + m\)

Step 10 ：\(m = \frac{4}{3} + \frac{1}{3} = \boxed{\frac{5}{3}}\)

Step 11 ：Now, we can find the value of \(x\) when \(f(x) = \frac{4}{3}\) using the value of \(m\) we found:

Step 12 ：\(\frac{4}{3} = 3x^2 - 2x + \frac{5}{3}\)

Step 13 ：\(0 = 3x^2 - 2x + \frac{5}{3} - \frac{4}{3}\)

Step 14 ：\(0 = 3x^2 - 2x + \frac{1}{3}\)

Step 15 ：To solve for \(x\), we can use the quadratic formula:

Step 16 ：\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Step 17 ：In this case, \(a = 3\), \(b = -2\), and \(c = \frac{1}{3}\):

Step 18 ：\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)\left(\frac{1}{3}\right)}}{2(3)}\)

Step 19 ：\(x = \frac{2 \pm \sqrt{4 - 4}}{6}\)

Step 20 ：\(x = \frac{2 \pm \sqrt{0}}{6}\)

Step 21 ：Since the square root of 0 is 0, we have:

Step 22 ：\(x = \frac{2}{6} = \boxed{\frac{1}{3}}\)