Question

Let \(f(x)=4 \sqrt{x}-x\). a. Find all points on the graph of \(f\) at which the tangent line is horizontal. b. Find all points on the graph of \(f\) at which the tangent line has slope \(-\frac{1}{2}\).

Short answer

Answer: The tangent line is horizontal at the point (4,4) and has a slope of -1/2 at the point (16,0).

Step-by-step solution

Find the derivative of the function

In order to find the points where the tangent line has specific slopes, we need to find the derivative of the function. To find the derivative of \(f(x)=4 \sqrt{x}-x\), we will use the power rule, sum and difference rule and the chain rule. The function can be rewritten as \(f(x)=4 x^{\frac{1}{2}}-x\). So, the derivative of the function is: $$f'(x) = \frac{d}{dx}(4x^{\frac{1}{2}}) - \frac{d}{dx}(x) = 4\cdot \frac{1}{2} x^{\frac{1}{2}-1} -1 = 2x^{-\frac{1}{2}} -1 $$

Find the points where the tangent line is horizontal

The tangent line is horizontal when the derivative of the function is equal to 0. So, we set the derivative equal to 0 and solve for \(x\): $$2x^{-\frac{1}{2}} -1 = 0$$ This gives us: $$x^{-\frac{1}{2}} = \frac{1}{2}$$ Next, we raise both sides to the power of -2 to isolate the variable x and get: $$x= (\frac{1}{2})^{-2} = 4$$ So, the point on the graph where the tangent line is horizontal is when \(x=4\). Then we find the corresponding \(y\)-value: $$y = f(4) = 4 \sqrt{4} - 4 = 4(2) - 4 = 4$$ Thus, the point on the graph where the tangent line is horizontal is \((4,4)\).

Find the points where the tangent line has a slope of \(-\frac{1}{2}\)

The tangent line has a slope of \(-\frac{1}{2}\) when the derivative of the function is equal to \(-\frac{1}{2}\). So, we set the derivative equal to \(-\frac{1}{2}\) and solve for \(x\): $$2x^{-\frac{1}{2}} -1 = -\frac{1}{2}$$ This gives us: $$x^{-\frac{1}{2}} = \frac{1}{4}$$ Next, we raise both sides to the power of -2 to isolate the variable x and get: $$x= (\frac{1}{4})^{-2} = 16$$ So, the point on the graph where the tangent line has a slope of \(-\frac{1}{2}\) is when \(x=16\). Then we find the corresponding \(y\)-value: $$y = f(16) = 4 \sqrt{16} - 16 = 4(4) - 16 = 0$$ Thus, the point on the graph where the tangent line has a slope of \(-\frac{1}{2}\) is \((16,0)\). In conclusion, the tangent line is horizontal at the point \((4,4)\) and has a slope of \(-\frac{1}{2}\) at the point \((16,0)\).

Key concepts

Derivative

The derivative of a function is a fundamental concept in calculus. It represents the rate of change of a function's output value with respect to changes in its input value. In simple terms, the derivative tells us how much a function is changing at any given point. For the function given in the exercise, which is \(f(x) = 4 \sqrt{x} - x\), we found its derivative using the power rule, the sum and difference rule, and the chain rule. These tools help break down complex expressions into manageable pieces for differentiation. The derivative obtained, \(f'(x) = 2x^{-\frac{1}{2}} - 1\), will guide us in understanding how the graph of the function behaves at different points.

Tangent Line

A tangent line to a curve at a particular point is a straight line that just "touches" the curve at that point. This line has the same slope as the curve itself at the point of contact, which means it mirrors the curve’s rate of change at that instant. Tangents are crucial in calculus because they provide linear approximations for curves, helping us analyze and estimate values of functions near certain points. In the context of this exercise, finding the tangent line involves using the derivative of the function to determine the slope at specific points, allowing us to characterize the behavior of the function at those points where the tangent line is either horizontal or has a specific slope.

Slope

The slope of a line is a measure of its steepness and direction. In calculus, the slope of the tangent line to a curve at a particular point is given by the derivative of the function at that point. The slope is positive if the tangent line rises as you move from left to right, negative if it falls, and zero if the line is horizontal. For the function \(f(x) = 4 \sqrt{x} - x\), the slope of the tangent line equals \(f'(x) = 2x^{-\frac{1}{2}} - 1\). This tells us how abruptly the curve turns and at what rate. It's the key to solving problems like determining where the tangent line is horizontal, which occurs when the slope equals zero, or where the tangent line matches a specific slope like \(-\frac{1}{2}\), as shown in the solution.

Horizontal tangent

A horizontal tangent occurs at a point on a curve where the slope is zero. This is because a horizontal line has no steepness, hence its slope is zero. Finding a horizontal tangent involves setting the derivative of the function equal to zero and solving for the input value, which gives us critical points—points of interest where the behavior of the function might change. In the given exercise, it was found that the derivative \(2x^{-\frac{1}{2}} - 1\) equals zero when \(x = 4\), indicating that at this point the tangent line to the graph of \(f(x)\) is horizontal. This understanding helps in analyzing and predicting the overall behavior of the function and its graph.