1. **State the problem:** Find the global maximum and minimum values of the function $$f(x) = 2x^2 - 8x + 7$$ on the interval $$[-3, 3]$$. 2. **Recall the formula and rules:** To find global extrema on a closed interval, evaluate the function at critical points inside the interval and at the endpoints. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(2x^2 - 8x + 7) = 4x - 8$$. 4. **Find critical points:** Set $$f'(x) = 0$$: $$4x - 8 = 0 \implies x = 2$$. 5. **Evaluate $$f(x)$$ at critical points and endpoints:** - At $$x = -3$$: $$f(-3) = 2(-3)^2 - 8(-3) + 7 = 18 + 24 + 7 = 49$$. - At $$x = 2$$: $$f(2) = 2(2)^2 - 8(2) + 7 = 8 - 16 + 7 = -1$$. - At $$x = 3$$: $$f(3) = 2(3)^2 - 8(3) + 7 = 18 - 24 + 7 = 1$$. 6. **Determine global max and min:** - Global maximum is $$49$$ at $$x = -3$$. - Global minimum is $$-1$$ at $$x = 2$$. **Final answer:** - Global maximum: $$49$$ at $$x = -3$$. - Global minimum: $$-1$$ at $$x = 2$$.