Questions: Use a graphing utility to approximate the local maximum value and local minimum value of the function f(x)=-0.3x^3-0.7x^2+4x-4 for -6<x<4. Using a graphing utility, graph the function for -6<x<4 and -25<y<10. Choose the correct graph below.

Solution Steps

The function given is \( f(x) = -0.3x^3 - 0.7x^2 + 4x - 4 \).

The range for \( x \) is \(-6 < x < 4\) and for \( y \) is \(-25 < y < 10\).

To find the local maximum and minimum, we need to find the derivative of the function and solve for critical points.

The derivative of \( f(x) \) is: \[ f'(x) = -0.9x^2 - 1.4x + 4 \]

Set the derivative equal to zero to find critical points: \[ -0.9x^2 - 1.4x + 4 = 0 \]

Using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = -0.9 \), \( b = -1.4 \), and \( c = 4 \).

Calculate the discriminant: \[ b^2 - 4ac = (-1.4)^2 - 4(-0.9)(4) = 1.96 + 14.4 = 16.36 \]

Calculate the roots: \[ x = \frac{-(-1.4) \pm \sqrt{16.36}}{2(-0.9)} = \frac{1.4 \pm 4.045}{-1.8} \]

The roots are: \[ x_1 = \frac{1.4 + 4.045}{-1.8} \approx -3.025 \] \[ x_2 = \frac{1.4 - 4.045}{-1.8} \approx 1.47 \]

Evaluate \( f(x) \) at \( x_1 \) and \( x_2 \) to find the local extrema.

For \( x_1 \approx -3.025 \): \[ f(-3.025) \approx -0.3(-3.025)^3 - 0.7(-3.025)^2 + 4(-3.025) - 4 \]

For \( x_2 \approx 1.47 \): \[ f(1.47) \approx -0.3(1.47)^3 - 0.7(1.47)^2 + 4(1.47) - 4 \]

Final Answer