A value of c for which the minimum value of f(x) = x² – 4cx + 8c is greater than the maximum value of g(x) = -x² + 3cx – 2c, is

The correct answer is the option represented by the provided image.

To find the correct value of c, we first need to determine the minimum value of the quadratic function f(x) and the maximum value of the quadratic function g(x).

The first function is given as:

$f(x)=x^2-4cx+8c$

Since the coefficient of $x^2$ is positive, this parabola opens upwards, meaning its vertex represents the minimum value. We can find the x-coordinate of the vertex using the formula $x=-\frac{b}{2a}$:

$x=\frac{-(-4c)}{2(1)}=2c$

Substitute this value back into f(x) to find the minimum value:

$f_{min}=(2c{)}^2-4c(2c)+8c=4c^2-8c^2+8c=-4c^2+8c$

The second function is given as:

$g(x)=-x^2+3cx-2c$

Since the coefficient of $x^2$ is negative, this parabola opens downwards, so its vertex represents the maximum value. The x-coordinate of the vertex is:

$x=\frac{-3c}{2(-1)}=\frac{3c}{2}$

Substitute this x value into g(x) to find the maximum value:

$g_{max}=-{\left(\frac{3c}{2}\right)}^2+3c\left(\frac{3c}{2}\right)-2c$

$g_{max}=-\frac{9c^2}{4}+\frac{9c^2}{2}-2c=\frac{9c^2}{4}-2c$

According to the problem, the minimum value of f(x) is greater than the maximum value of g(x). Therefore, we can set up the following inequality:

$-4c^2+8c>\frac{9c^2}{4}-2c$

To clear the fraction, multiply the entire inequality by 4:

$-16c^2+32c>9c^2-8c$

Now, bring all the terms to one side of the inequality to form a quadratic inequality:

$25c^2-40c<0$

Factor out the common term, which is 5c:

$5c(5c-8)<0$

This product is strictly less than 0 when c lies between the roots of the quadratic equation. The roots are 0 and 8/5. Therefore, the valid range for the parameter c is:

$c\in (0,\frac{8}{5})$

Any value within this open interval satisfies the condition. The correct provided image option represents a value of c that falls in the interval (0, 8/5).