91Ó°ÊÓ

Step 1: Compute the derivative of f(x). We are given \(f(x)=4\sqrt{x}-x\). Let's find the derivative of function f(x). #tag_formula#$$\frac{d}{dx} [4\sqrt{x}-x]=\frac{d}{dx} [4 x^{\frac{1}{2}} - x] = 4\cdot \frac{1}{2} x^{-\frac{1}{2}} -1 = 2 x^{-\frac{1}{2}} -1 .$$#tag_formula_end# Step 2: Find the x-values where f'(x) = 0 (horizontal tangent). Next, let's find the x-values at which the derivative of the function is 0, as the tangent line will be horizontal when f'(x)=0: #tag_formula#$$2 x^{-\frac{1}{2}} -1 = 0.$$ #tag_formula_end# Solving the above equation for x, we get: #tag_formula#$$2 x^{-\frac{1}{2}} = 1.$$#tag_formula_end# So, #tag_formula#$$x^{-\frac{1}{2}} = \frac{1}{2}.$$#tag_formula_end# From this, we find the value of x: #tag_formula#$$x = \left(\frac{1}{2}\right)^{-2} = 4.$$#tag_formula_end# Step 3: Calculate the corresponding y-value for the point. Now, to find the coordinate of the point on the graph, we need to compute f(x) at x = 4: #tag_formula#$$f(4) = 4\sqrt{4} - 4 = 8 - 4 = 4.$$#tag_formula_end# Therefore, the point where the tangent line is horizontal is (4, 4).

Step 1: Determine where f'(x) equals \(-\frac{1}{2}\) Using the derivative calculated in part (a), we now need to find the x-values at which the slope of the tangent line is \(-\frac{1}{2}\): #tag_formula#$$2 x^{-\frac{1}{2}} - 1 = -\frac{1}{2}.$$ #tag_formula_end# Step 2: Solve for x. Now, solving this equation for x, we get: #tag_formula#$$2 x^{-\frac{1}{2}} = \frac{1}{2}.$$#tag_formula_end# So, #tag_formula#$$x^{-\frac{1}{2}} = \frac{1}{4}.$$#tag_formula_end# From this, we find the value of x: #tag_formula#$$x = \left(\frac{1}{4}\right)^{-2} = 16.$$#tag_formula_end# Step 3: Calculate the corresponding y-value for the point. Finally, to find the coordinate of the point on the graph, we need to compute f(x) at x = 16: #tag_formula#$$f(16) = 4\sqrt{16} - 16 = 16 - 16 = 0.$$#tag_formula_end# Thus, the point where the tangent line has a slope of \(-\frac{1}{2}\) is (16, 0).