Problem 56
Question
Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. $$ y=x^{2}+2 x $$
Step-by-Step Solution
Verified Answer
The function \(y = x^{2} + 2x\) has a horizontal tangent at the point (-1, -1).
1Step 1: Find the derivative of the function
The derivative of \(y = x^{2} + 2x\) is given by applying the power rule of derivatives, i.e., if we have \(y = ax^{n}\), then its derivative \(\(y'= n*ax^{n-1}\) \). So, \(y' = 2x + 2\)
2Step 2: Solve for x where the derivative equals zero
Setting the derivative equal to zero, we solve for x to find the x-coordinate(s) of the point(s) where the function has a horizontal tangent line. 2x + 2 = 0. Solving this equation for x gives x = -1.
3Step 3: Substitute x into the original function
To get the y-coordinate of the point where the slope is zero, we substitute the obtained x-value into the original function. Substituting x = -1 into the original function \(y = x^{2} + 2x\), we get \(y = (-1)^{2} + 2(-1) = -1\). Therefore, the function \(y = x^{2} + 2x\) has a horizontal tangent at the point (-1, -1).
Key Concepts
DerivativePower RuleFunction GraphSlope
Derivative
When working with derivatives, we are essentially finding the rate at which a function's value changes at any given point. For a function like \(y = x^{2} + 2x\), the derivative helps us understand how the slope of the function graph changes. To find the derivative, we use the power rule, which simplifies the process of differentiation.
Power Rule
The power rule is a quick way to differentiate functions that have variables raised to a power. If you have a term like \(ax^n\), where \(a\) is a constant and \(n\) is the exponent, the derivative is \(n \cdot ax^{n-1}\). For the function \(y = x^{2} + 2x\), applying the power rule means:
- The derivative of \(x^2\) is \(2x^{2-1} = 2x\).
- The derivative of \(2x\) is simply \(2\) because the first power drops out, as \(x^1\) becomes \(x^0 = 1\).
Function Graph
A function graph is a visual representation of all the possible points \((x, y)\) that satisfy a function equation. For \(y = x^{2} + 2x\), the graph is a parabola. Understanding how this graph looks helps us find features like intercepts and tangent lines.
The slope of the tangent line changes at different points on the graph. By analyzing these, we pinpoint where the slope is zero, which indicates horizontal tangent lines, providing us information about key points and changes in direction.
Graphs allow us to visually interpret the derivative's results. They show us what happens when we solve for slopes, especially as they approach zero, connecting concepts to real geometric interpretations on the graph.
The slope of the tangent line changes at different points on the graph. By analyzing these, we pinpoint where the slope is zero, which indicates horizontal tangent lines, providing us information about key points and changes in direction.
Graphs allow us to visually interpret the derivative's results. They show us what happens when we solve for slopes, especially as they approach zero, connecting concepts to real geometric interpretations on the graph.
Slope
Slope is a measure of steepness or the inclination of a line. For functions, the slope tells us how fast they are changing at any point. Calculating the slope of tangent lines involves finding the derivative at specific \(x\)-values.
To determine where a function has a horizontal tangent line, we set its slope (or derivative) to zero and solve for \(x\). In our equations, a horizontal tangent means a zero slope. This equates to setting \(y' = 2x + 2 = 0\). Here the solution, \(x = -1\), informs us that the slope is zero at this point on the function.
By plugging \(x = -1\) back into the original equation \(y = x^{2} + 2x\), we find \(y = -1\), confirming that at the point \((-1, -1)\), the slope is indeed zero, thus confirming a horizontal tangent line.
To determine where a function has a horizontal tangent line, we set its slope (or derivative) to zero and solve for \(x\). In our equations, a horizontal tangent means a zero slope. This equates to setting \(y' = 2x + 2 = 0\). Here the solution, \(x = -1\), informs us that the slope is zero at this point on the function.
By plugging \(x = -1\) back into the original equation \(y = x^{2} + 2x\), we find \(y = -1\), confirming that at the point \((-1, -1)\), the slope is indeed zero, thus confirming a horizontal tangent line.
Other exercises in this chapter
Problem 55
Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=\sqrt{x-1} $$
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Use the demand function to find the rate of change in the demand \(x\) for the given price \(p .\) $$ x=275\left(1-\frac{3 p}{5 p+1}\right), p=\$ 4 $$
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determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The second derivative represents the ra
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Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=\frac{1}{\sqrt{x+2}} $$
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