Problem 45
Question
Let \(f(x)=x^{3}\). a. Find the point on the graph of \(f\) where the tangent line is horizontal. b. Sketch the graph of \(f\) and draw the horizontal tangent line.
Step-by-Step Solution
Verified Answer
The point on the graph of \(f(x) = x^3\) with a horizontal tangent line is (0, 0). To find this, we calculated the derivative \(f'(x) = 3x^2\), set it equal to 0, and solved for x, resulting in \(x = 0\). We then substituted this x value back into the original function to find the corresponding y value, resulting in the point (0, 0). Sketching the graph, we see the tangent line coincides with the x-axis at this point.
1Step 1: Compute the derivative of the function.
Calculate the derivative of the function \( f(x) = x^3 \) using differentiation rule (power rule):
\( f'(x) = \frac{d}{dx}(x^3) = 3x^2 \)
2Step 2: Find the value of x where the derivative is equal to zero.
Set the derivative equal to zero and solve for x:
\( 3x^2 = 0 \)
Divide both sides by 3:
\( x^2 = 0 \)
Take the square root of both sides:
\( x = 0 \)
3Step 3: Determine the point on the curve that corresponds to the found x value.
Plug the found x value into the original function to find the corresponding y value:
\( f(0) = (0)^3 = 0 \)
So the point on the graph where the tangent line is horizontal is (0, 0).
4Step 4: Draw the graph of f(x) and the tangent line.
Sketch the graph of the cubic function \( f(x) = x^3 \) and draw the horizontal tangent line at the point (0, 0). You'll notice the function has a smooth curve passing through the origin, and the tangent line is a horizontal line (the x-axis) that touches the curve only at the point (0, 0).
Key Concepts
Understanding the Power RuleCalculating the DerivativeIdentifying the Horizontal Tangent Line
Understanding the Power Rule
The power rule is a fundamental concept in differential calculus that simplifies the process of finding derivatives. It is particularly useful when dealing with polynomial functions. The power rule states that if you have a function of the form \( f(x) = x^n \), its derivative is \( f'(x) = nx^{n-1} \). This means you multiply the exponent \( n \) by the function and reduce the exponent by one.
For example, given the function \( f(x) = x^3 \), applying the power rule yields the derivative: \( f'(x) = 3x^{3-1} = 3x^2 \).
With the power rule, you can quickly and easily find the rate of change of a polynomial function, which is essential in understanding various calculus concepts such as tangent lines and optimization.
For example, given the function \( f(x) = x^3 \), applying the power rule yields the derivative: \( f'(x) = 3x^{3-1} = 3x^2 \).
With the power rule, you can quickly and easily find the rate of change of a polynomial function, which is essential in understanding various calculus concepts such as tangent lines and optimization.
Calculating the Derivative
Once you understand the power rule, calculating derivatives becomes straightforward. The derivative function describes how the original function \( f(x) \) changes at any point \( x \). For the function \( f(x) = x^3 \), using the power rule, we've found the derivative to be \( f'(x) = 3x^2 \).
This derivative function \( f'(x) \) is crucial because it allows us to determine the slope of the tangent line at any point on the curve of the original function. For any value of \( x \), plugging it into \( f'(x) \) gives the slope of the curve at that point.
The process of finding this derivative involves recognizing the function form, applying the power rule correctly, and simplifying the result, usually only requiring a few straightforward algebraic steps.
This derivative function \( f'(x) \) is crucial because it allows us to determine the slope of the tangent line at any point on the curve of the original function. For any value of \( x \), plugging it into \( f'(x) \) gives the slope of the curve at that point.
The process of finding this derivative involves recognizing the function form, applying the power rule correctly, and simplifying the result, usually only requiring a few straightforward algebraic steps.
Identifying the Horizontal Tangent Line
A horizontal tangent line on the curve of \( f(x) = x^3 \) indicates where the slope of the curve is zero. A tangent line is horizontal when its slope is zero, which corresponds to when the derivative of the function equals zero.
For the derivative \( f'(x) = 3x^2 \), setting it equal to zero gives \( 3x^2 = 0 \). Solving for \( x \), we divide by 3 to get \( x^2 = 0 \), and take the square root of both sides, resulting in \( x = 0 \).
At \( x = 0 \), the slope of the tangent line is zero, meaning the tangent is horizontal. To find the exact point on the curve, substitute \( x = 0 \) into the original function \( f(x) = x^3 \), yielding \( f(0) = 0 \). Therefore, the point is \((0, 0)\).
Understanding horizontal tangent lines is crucial in differential calculus as they can signify critical points or turning points in a graph. They provide insights into the behavior of functions at specific points.
For the derivative \( f'(x) = 3x^2 \), setting it equal to zero gives \( 3x^2 = 0 \). Solving for \( x \), we divide by 3 to get \( x^2 = 0 \), and take the square root of both sides, resulting in \( x = 0 \).
At \( x = 0 \), the slope of the tangent line is zero, meaning the tangent is horizontal. To find the exact point on the curve, substitute \( x = 0 \) into the original function \( f(x) = x^3 \), yielding \( f(0) = 0 \). Therefore, the point is \((0, 0)\).
Understanding horizontal tangent lines is crucial in differential calculus as they can signify critical points or turning points in a graph. They provide insights into the behavior of functions at specific points.
Other exercises in this chapter
Problem 45
Find the derivative of the function. \(g(t)=\frac{\sqrt{t+1}}{\sqrt{t^{2}+1}}\)
View solution Problem 45
Find the first and second derivatives of the given function. \(f(x)=2 x^{3}-3 x^{2}+1\)
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Let \(x\) and \(f(x)\) represent the given quantities. Fix \(x=a\) and let \(h\) be a small positive number. Give an interpretation of the quantities $$ \frac{f
View solution Problem 45
In Exercises 45-56, find the values of \(x\) for which each function is continuous. \(f(x)=2 x^{2}+x-1\)
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