1) Find the maximum or mintumam value of (a) \( 4<1-4 x-6 x^{2} \) (b) \( 4=4 x^{2}-5 x+1 \)

To find the maximum or minimum values of the given expressions, we will analyze each one separately. ### (a) \( 4 < 1 - 4x - 6x^2 \) First, we can rearrange the inequality: \[ 1 - 4x - 6x^2 - 4 > 0 \] This simplifies to: \[ -6x^2 - 4x - 3 > 0 \] To find the critical points, we can set the left-hand side to zero: \[ -6x^2 - 4x - 3 = 0 \] We can multiply through by -1 to make the leading coefficient positive: \[ 6x^2 + 4x + 3 = 0 \] Next, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): Here, \( a = 6 \), \( b = 4 \), and \( c = 3 \). Calculating the discriminant: \[ b^2 - 4ac = 4^2 - 4 \cdot 6 \cdot 3 = 16 - 72 = -56 \] Since the discriminant is negative, there are no real roots, which means the quadratic does not cross the x-axis. Since the leading coefficient (6) is positive, the quadratic opens upwards and is always positive. Thus, the inequality \( -6x^2 - 4x - 3 > 0 \) holds for all \( x \). Therefore, there is no maximum or minimum value in the traditional sense, as the expression is always greater than 4. ### (b) \( 4 = 4x^2 - 5x + 1 \) We can rearrange this equation to find the values of \( x \): \[ 4x^2 - 5x + 1 - 4 = 0 \] This simplifies to: \[ 4x^2 - 5x - 3 = 0 \] Now we can use the quadratic formula again: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 4 \), \( b = -5 \), and \( c = -3 \). Calculating the discriminant: \[ b^2 - 4ac = (-5)^2 - 4 \cdot 4 \cdot (-3) = 25 + 48 = 73 \] Since the discriminant is positive, there are two real roots: \[ x = \frac{5 \pm \sqrt{73}}{8} \] These roots represent the points where the quadratic equals 4. To find the maximum or minimum value of the quadratic \( 4x^2 - 5x + 1 \), we can find the vertex of the parabola. The x-coordinate of the vertex is given by: \[ x = -\frac{b}{2a} = -\frac{-5}{2 \cdot 4} = \frac{5}{8} \] Now we can substitute \( x = \frac{5}{8} \) back into the quadratic to find the minimum value: \[ 4\left(\frac{5}{8}\right)^2 - 5\left(\frac{5}{8}\right) + 1 \] Calculating this: \[ = 4 \cdot \frac{25}{64} - \frac{25}{8} + 1 \] \[ = \frac{100}{64} - \frac{200}{64} + \frac{64}{64} \] \[ = \frac{100 - 200 + 64}{64} = \frac{-36}{64} = -\frac{9}{16} \] Thus, the minimum value of \( 4x^2 - 5x + 1 \) is \( -\frac{9}{16} \). ### Summary: - For (a), the expression \( 4 < 1 - 4x - 6x^2 \) is always true for all \( x \). - For (b), the minimum value of \( 4x^2 - 5x + 1 \) is \( -\frac{9}{16} \).