1. **State the problem:** We have the function $$f(x) = 2(x^2 - 3)^2$$ and need to solve parts c and d. 2. **Part c: If $$f(x) = 6$$, find possible values of $$x$$.** Start with the equation: $$2(x^2 - 3)^2 = 6$$ Divide both sides by 2: $$\cancel{2}(x^2 - 3)^2 = \cancel{2}3$$ $$ (x^2 - 3)^2 = 3 $$ Take the square root of both sides: $$ x^2 - 3 = \pm \sqrt{3} $$ Solve for $$x^2$$: $$ x^2 = 3 \pm \sqrt{3} $$ This gives two cases: - $$x^2 = 3 + \sqrt{3}$$ - $$x^2 = 3 - \sqrt{3}$$ Find $$x$$ by taking square roots: $$ x = \pm \sqrt{3 + \sqrt{3}} $$ $$ x = \pm \sqrt{3 - \sqrt{3}} $$ So the possible values of $$x$$ are: $$ \pm \sqrt{3 + \sqrt{3}}, \pm \sqrt{3 - \sqrt{3}} $$ 3. **Part d: Does $$f(3) = 4$$ for this function? Explain.** Calculate $$f(3)$$: $$ f(3) = 2(3^2 - 3)^2 = 2(9 - 3)^2 = 2(6)^2 = 2 \times 36 = 72 $$ Since $$f(3) = 72$$ and not 4, the statement $$f(3) = 4$$ is false. **Summary:** - For $$f(x) = 6$$, $$x = \pm \sqrt{3 + \sqrt{3}}$$ or $$x = \pm \sqrt{3 - \sqrt{3}}$$. - $$f(3)$$ equals 72, so $$f(3) = 4$$ is incorrect.