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Magazine Chapter 3: Problem 56
Find the point \(P\) on the graph of \(y=x^{3}\) such that the tangent line at \(P\) has \(x\) -intercept 4.
Short Answer
Expert verified
The point P is (6, 216).
Step by step solution
01
Understand the Tangent Line Equation
The equation of a tangent line to the curve at point \((x_0, y_0)\) is \(y - y_0 = m(x - x_0)\), where \(m\) is the derivative of the function at \(x_0\), i.e., \(m = f'(x_0)\). In our case, \(f(x) = x^3\), so we need to find its derivative \(f'(x) = 3x^2\).
02
Point on the Graph
Let the point \(P\) be \((a, a^3)\). We need to determine the value of \(a\) such that the tangent line at point \((a, a^3)\) has an \(x\)-intercept of 4. The slope of the tangent line at this point is given by \(f'(a) = 3a^2\).
03
Form Tangent Equation with Given Condition
The tangent line at \((a, a^3)\) is expressed as \(y - a^3 = 3a^2(x - a)\). Since the \(x\)-intercept of the tangent line is 4, substitute \(y = 0\) in the line equation to find when the line crosses the x-axis: \[0 - a^3 = 3a^2(4 - a)\]yields the equation for the intercept.
04
Solve for 'a'
From \(-a^3 = 3a^2(4 - a)\), simplify to get \[-a^3 = 12a^2 - 3a^3\]Rearrange to:\[2a^3 = 12a^2\]Divide both sides by \(2a^2\) (assuming \(a eq 0\)):\[a = 6\]
05
Verification
Verify by substituting \(a = 6\) back into the equations. The point \(P\) is \((6, 6^3) = (6, 216)\). The tangent line at \(x = 6\) is \[y - 216 = 3(6)^2(x - 6)\]or\[y - 216 = 108(x - 6)\]Simplifying gives the equation \[y = 108x - 432\].For the \(x\)-intercept, set \(y = 0\):\[0 = 108x - 432\] which results in \(x = 4\). This confirms the computations.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivative
The derivative is a core concept in calculus that measures how a function changes as its input changes. It represents the slope or steepness of the tangent line at a given point on a curve.
For the function given in the problem, which is \( f(x) = x^3 \), the derivative is calculated using the power rule. The power rule states that if \( f(x) = x^n \), then the derivative \( f'(x) = n \times x^{n-1} \).
Applying this rule here, we find the derivative of \( x^3 \) to be \( 3x^2 \).
For the function given in the problem, which is \( f(x) = x^3 \), the derivative is calculated using the power rule. The power rule states that if \( f(x) = x^n \), then the derivative \( f'(x) = n \times x^{n-1} \).
Applying this rule here, we find the derivative of \( x^3 \) to be \( 3x^2 \).
- The derivative tells us how the function \( y = x^3 \) changes at any point \( x_0 \) on the graph.
- The derivative at a point also determines the slope of the tangent line at that point.
Polynomial Function
A polynomial function is a type of mathematical expression involving a sum of powers in one or more variables multiplied by coefficients. In this problem, the polynomial function in question is \( f(x) = x^3 \).
This particular function is a cubic polynomial, which means it has a degree of 3. The degree of a polynomial is determined by the highest power of the variable present in the function.
This particular function is a cubic polynomial, which means it has a degree of 3. The degree of a polynomial is determined by the highest power of the variable present in the function.
- Given its cubic nature, \( f(x) = x^3 \) will have one real root and can change its behavior through turning points where the slope of the function is zero.
- Polynomial functions like \( x^3 \) are smooth and continuous, making calculations like derivatives straightforward.
X-intercept
The x-intercept of a line is the point where the line crosses the x-axis. At this point, the y-coordinate is always zero.
For this exercise, the requirement is for the tangent line at point \( P \) to have an x-intercept of 4. What this means is, when the tangent line equation is rearranged or solved for \( x \), where \( y = 0 \), the solution should be 4.
Finding the x-intercept involves substituting \( y = 0 \) into the equation of the tangent line:
For this exercise, the requirement is for the tangent line at point \( P \) to have an x-intercept of 4. What this means is, when the tangent line equation is rearranged or solved for \( x \), where \( y = 0 \), the solution should be 4.
Finding the x-intercept involves substituting \( y = 0 \) into the equation of the tangent line:
- This equation derived from the problem was \( y - a^3 = 3a^2(x - a) \).
- Setting \( y = 0 \), and solving for \( x \) provides the desired x-intercept.
Slope
Slope is a measure of how much a line ascends or descends. For a tangent line, the slope is determined by the derivative at a particular point on the curve.
In linear equations, slope is frequently represented by m in the formula \( y - y_0 = m(x - x_0) \).
In this context, the slope specifies how fast the function's value is changing at any given point.
In linear equations, slope is frequently represented by m in the formula \( y - y_0 = m(x - x_0) \).
In this context, the slope specifies how fast the function's value is changing at any given point.
- The slope of the tangent line at point \( P(a, a^3) \) is \( m = 3a^2 \), derived from the function's derivative \( 3x^2 \).
- An understanding of slope is essential for constructing the tangent line, as it directly influences its angle and direction.
