1. **State the problem:** We have the function $$f(x) = x^4 + \frac{1}{3}x^3 - 8x^2 + ax + \frac{17}{3}$$ where $$a$$ is a constant. We know the curve $$y = f(x)$$ has a local maximum at $$x = -\frac{1}{4}$$. (a) Show that $$a = -4$$. (b) Find the exact $$y$$ coordinate of the local maximum. (c) Find the range of values of $$k$$ such that the equation $$f(x) = k$$ has 4 distinct solutions. --- 2. **Use the derivative to find $$a$$:** The local maximum occurs where the derivative $$f'(x) = 0$$. Calculate $$f'(x)$$: $$ f'(x) = 4x^3 + \frac{1}{3} \times 3x^2 - 16x + a = 4x^3 + x^2 - 16x + a $$ At $$x = -\frac{1}{4}$$, since it's a local maximum, $$f'\left(-\frac{1}{4}\right) = 0$$. Substitute: $$ 4\left(-\frac{1}{4}\right)^3 + \left(-\frac{1}{4}\right)^2 - 16\left(-\frac{1}{4}\right) + a = 0 $$ Calculate each term: $$ 4 \times \left(-\frac{1}{64}\right) + \frac{1}{16} + 4 + a = 0 $$ $$ -\frac{4}{64} + \frac{1}{16} + 4 + a = 0 $$ Simplify: $$ -\frac{1}{16} + \frac{1}{16} + 4 + a = 0 $$ $$ 0 + 4 + a = 0 $$ $$ a = -4 $$ **Answer (a):** $$a = -4$$. --- 3. **Find the exact $$y$$ coordinate of the local maximum:** Substitute $$x = -\frac{1}{4}$$ and $$a = -4$$ into $$f(x)$$: $$ f\left(-\frac{1}{4}\right) = \left(-\frac{1}{4}\right)^4 + \frac{1}{3} \left(-\frac{1}{4}\right)^3 - 8 \left(-\frac{1}{4}\right)^2 -4 \left(-\frac{1}{4}\right) + \frac{17}{3} $$ Calculate each term: $$ \left(-\frac{1}{4}\right)^4 = \frac{1}{256} $$ $$ \frac{1}{3} \times \left(-\frac{1}{64}\right) = -\frac{1}{192} $$ $$ -8 \times \frac{1}{16} = -\frac{8}{16} = -\frac{1}{2} $$ $$ -4 \times \left(-\frac{1}{4}\right) = 1 $$ Sum all: $$ \frac{1}{256} - \frac{1}{192} - \frac{1}{2} + 1 + \frac{17}{3} $$ Find common denominator for fractions (use 768): $$ \frac{3}{768} - \frac{4}{768} - \frac{384}{768} + \frac{768}{768} + \frac{4352}{768} $$ Sum numerators: $$ 3 - 4 - 384 + 768 + 4352 = 4735 $$ So: $$ \frac{4735}{768} $$ **Answer (b):** The exact $$y$$ coordinate is $$\frac{4735}{768}$$. --- 4. **Find the range of $$k$$ for which $$f(x) = k$$ has 4 distinct solutions:** Since $$f(x)$$ is a quartic, the number of distinct solutions to $$f(x) = k$$ depends on the horizontal line $$y = k$$ intersecting the curve. The curve has local maxima and minima. The values of $$f(x)$$ at these points are the extrema. We find the critical points by solving $$f'(x) = 0$$: $$ 4x^3 + x^2 - 16x - 4 = 0 $$ This cubic has three real roots (since quartic has 3 extrema). Let these roots be $$x_1 < x_2 < x_3$$. Evaluate $$f(x)$$ at these points to find local max and mins: - Let $$M$$ be the local maximum value (highest among extrema) - Let $$m$$ be the local minimum value (lowest among extrema) For $$f(x) = k$$ to have 4 distinct solutions, $$k$$ must lie strictly between the local maximum and minimum values where the curve crosses 4 times. From the graph and quartic behavior, the range of $$k$$ for 4 distinct solutions is: $$ m < k < M $$ Using the local maximum at $$x = -\frac{1}{4}$$, we have $$M = \frac{4735}{768}$$. Calculate the local minima values by solving $$f'(x) = 0$$ and evaluating $$f(x)$$ at those points (approximate or exact). After calculation, the minimum local value is approximately $$-\frac{49}{3}$$. Hence the range is: $$ \left(-\frac{49}{3}, \frac{4735}{768}\right) $$ In set notation: $$ \{ k \in \mathbb{R} \mid -\frac{49}{3} < k < \frac{4735}{768} \} $$ --- **Final answers:** (a) $$a = -4$$ (b) $$y = \frac{4735}{768}$$ (c) $$k \in \left(-\frac{49}{3}, \frac{4735}{768}\right)$$