1. **Problem Statement:** We are given the function $$f(x) = x^4 + \frac{1}{2}x^3 - 5x^2 + \tan\left(\frac{x}{2}\right)$$ and asked to find at which of the given points $$x = -2, -1, 0, 1$$ the tangent line to the graph of $$f$$ has the greatest slope. 2. **Understanding the problem:** The slope of the tangent line to the graph of $$f$$ at a point $$x$$ is given by the derivative $$f'(x)$$. To find where the tangent line has the greatest slope among the given points, we need to compute $$f'(x)$$ and evaluate it at each point. 3. **Find the derivative:** Using the sum rule and chain rule, $$ f'(x) = \frac{d}{dx}\left(x^4\right) + \frac{d}{dx}\left(\frac{1}{2}x^3\right) - \frac{d}{dx}\left(5x^2\right) + \frac{d}{dx}\left(\tan\left(\frac{x}{2}\right)\right) $$ Calculate each term: - $$\frac{d}{dx} x^4 = 4x^3$$ - $$\frac{d}{dx} \frac{1}{2}x^3 = \frac{1}{2} \cdot 3x^2 = \frac{3}{2}x^2$$ - $$\frac{d}{dx} 5x^2 = 10x$$ - $$\frac{d}{dx} \tan\left(\frac{x}{2}\right) = \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} = \frac{1}{2} \sec^2\left(\frac{x}{2}\right)$$ So, $$ f'(x) = 4x^3 + \frac{3}{2}x^2 - 10x + \frac{1}{2} \sec^2\left(\frac{x}{2}\right) $$ 4. **Evaluate $$f'(x)$$ at each given point:** - At $$x = -2$$: $$ f'(-2) = 4(-2)^3 + \frac{3}{2}(-2)^2 - 10(-2) + \frac{1}{2} \sec^2\left(-1\right) = 4(-8) + \frac{3}{2}(4) + 20 + \frac{1}{2} \sec^2(-1) = -32 + 6 + 20 + \frac{1}{2} \sec^2(-1) = -6 + \frac{1}{2} \sec^2(-1) $$ Since $$\sec^2(-1) = \sec^2(1)$$ (because cosine is even), approximate $$\cos(1) \approx 0.5403$$, so $$\sec^2(1) = \frac{1}{\cos^2(1)} \approx \frac{1}{0.5403^2} = \frac{1}{0.292} \approx 3.424$$ Thus, $$f'(-2) \approx -6 + \frac{1}{2} \times 3.424 = -6 + 1.712 = -4.288$$ - At $$x = -1$$: $$ f'(-1) = 4(-1)^3 + \frac{3}{2}(-1)^2 - 10(-1) + \frac{1}{2} \sec^2\left(-\frac{1}{2}\right) = 4(-1) + \frac{3}{2}(1) + 10 + \frac{1}{2} \sec^2\left(-0.5\right) = -4 + 1.5 + 10 + \frac{1}{2} \sec^2(0.5) = 7.5 + \frac{1}{2} \sec^2(0.5) $$ Approximate $$\cos(0.5) \approx 0.8776$$, so $$\sec^2(0.5) = \frac{1}{0.8776^2} = \frac{1}{0.770} \approx 1.2987$$ Thus, $$f'(-1) \approx 7.5 + \frac{1}{2} \times 1.2987 = 7.5 + 0.649 = 8.149$$ - At $$x = 0$$: $$ f'(0) = 4(0)^3 + \frac{3}{2}(0)^2 - 10(0) + \frac{1}{2} \sec^2(0) = 0 + 0 + 0 + \frac{1}{2} \times 1 = \frac{1}{2} = 0.5 $$ - At $$x = 1$$: $$ f'(1) = 4(1)^3 + \frac{3}{2}(1)^2 - 10(1) + \frac{1}{2} \sec^2\left(\frac{1}{2}\right) = 4 + 1.5 - 10 + \frac{1}{2} \sec^2(0.5) = -4.5 + \frac{1}{2} \times 1.2987 = -4.5 + 0.649 = -3.851 $$ 5. **Compare the slopes:** - $$f'(-2) \approx -4.288$$ - $$f'(-1) \approx 8.149$$ - $$f'(0) = 0.5$$ - $$f'(1) \approx -3.851$$ The greatest slope is at $$x = -1$$ with approximately $$8.149$$. **Final answer:** The tangent line has the greatest slope at $$\boxed{-1}$$ (option B).