Question

\(f_{1}^{\prime}(x)=2 x-1 \quad \& \quad f_{2}^{\prime}(x)=3 x^{2}-2 x-2\) \(\Rightarrow 2 x_{1}-1=3 x_{2}^{2}-2 x_{2}-2\) \(\Rightarrow 3 x_{2}^{2}-2 x_{2}-2 x_{1}-1=0\) For \(\mathrm{x}_{2}\) be real, \(\mathrm{D} \geq 0\) \(4-4\left(2 x_{1}+1\right)(3)\) \(4-24 x_{1}-12\) \(\Rightarrow-\left(24 x_{1}+8\right)\) There can be infinite such values of \(\mathrm{x}_{\mathrm{l}}\)

Short answer

Answer: \(x_1 \geq -\frac{2}{3}\)

Step-by-step solution

Given in the problem, we have \(f_{1}^{\prime}(x_{1})=f_{2}^{\prime}(x_{2})\). Plugging in the given expressions for \(f_1'\) and \(f_2'\), we have: \(2 x_{1}-1=3 x_{2}^{2}-2 x_{2}-2\) #Step 2: Rearrange to get the equation in terms of \(x_2\)#

Rearranging the equation from Step 1, we can obtain the equation: \(3 x_{2}^{2}-2 x_{2}-2 x_{1}-1=0\) #Step 3: Write the discriminant expression for \(x_2\)#

In order to determine if the equation has real roots for \(x_2\), we must check if the discriminant (\(\Delta\)) is greater than or equal to zero. For a quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant \(\Delta\) is given by: \(\Delta = b^2 - 4ac\). In our case, \(a=3\), \(b=-2\), and \(c= -(2x_1 +1)\). Substituting these values, we get: \(\Delta = (-2)^2 - 4(3)(-2x_1 -1)\) #Step 4: Simplify the discriminant expression#

Simplifying the expression from Step 3, we have: \(\Delta = 4 + 24x_1 + 12\) #Step 5: Find the condition for which \(x_2\) has real roots#

For \(x_2\) to have real roots, the discriminant \(\Delta\) must be greater than or equal to zero. So we have the inequality: \(\Delta = 4 + 24x_1 + 12 \geq 0\) #Step 6: Simplify the inequality and find \(x_1\) condition#

Simplifying the inequality from Step 5, we obtain: \(24x_1 + 16 \geq 0\) Dividing by \(24\) to isolate the variable, we get: \(x_1 \geq -\frac{2}{3}\) It means that for any value of \(x_1\) greater than or equal to \(-\frac{2}{3}\), there are real values of \(x_2\).