Problem 21
Question
If \(f(x)=3 x^{2}-5 x,\) find \(f^{\prime}(2)\) and use it to find an equation of the tangent line to the parabola \(y=3 x^{2}-5 x\) at the point \((2,2) .\)
Step-by-Step Solution
Verified Answer
The tangent line at (2,2) is \(y = 7x - 12\).
1Step 1: Find the derivative of the function
The function given is \(f(x) = 3x^2 - 5x\). To find the derivative, we use the power rule. Differentiate each term separately: the derivative of \(3x^2\) is \(6x\), and the derivative of \(-5x\) is \(-5\). Thus, \(f'(x) = 6x - 5\).
2Step 2: Evaluate the derivative at x = 2
Now that we have the derivative \(f'(x) = 6x - 5\), we find \(f'(2)\) by substituting \(x = 2\) into the derivative. This gives us: \(f'(2) = 6(2) - 5 = 12 - 5 = 7\).
3Step 3: Confirm the point of tangency on the curve
Substitute \(x = 2\) into the original function to confirm the y-coordinate of the point. \(f(2) = 3(2)^2 - 5(2) = 12 - 10 = 2\), verifying that the point \((2,2)\) lies on the curve \(y = 3x^2 - 5x\).
4Step 4: Write the equation of the tangent line
The equation of the tangent line can be expressed in point-slope form: \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is the point of tangency \((2,2)\) and \(m\) is the slope, which is \(f'(2) = 7\). Substituting these values gives: \(y - 2 = 7(x - 2)\). Simplifying, we get \(y = 7x - 14 + 2 = 7x - 12\).
Key Concepts
Power RuleTangent LineDerivative Evaluation
Power Rule
The Power Rule is a foundational concept in calculus used to find the derivative of functions involving powers of a variable. If you have a function in the form of \(f(x) = ax^n\), the derivative \(f'(x)\) can be calculated by multiplying the exponent \(n\) by the coefficient \(a\) and then reducing the exponent by one. In mathematical terms, this is expressed as \(f'(x) = anx^{n-1}\).
For the function \(f(x) = 3x^2 - 5x\), we apply the power rule to differentiate each term separately. The term \(3x^2\) gives us the derivative \(6x\) and \(-5x\) gives the constant derivative \(-5\). Therefore, the derivative of the function is \(f'(x) = 6x - 5\).
Remember: The power rule is effective for any polynomial term and forms the basis for solving more complex differentiation problems. By mastering this rule, you'll find it easier to handle various calculus problems that involve finding slopes and rates of change.
For the function \(f(x) = 3x^2 - 5x\), we apply the power rule to differentiate each term separately. The term \(3x^2\) gives us the derivative \(6x\) and \(-5x\) gives the constant derivative \(-5\). Therefore, the derivative of the function is \(f'(x) = 6x - 5\).
Remember: The power rule is effective for any polynomial term and forms the basis for solving more complex differentiation problems. By mastering this rule, you'll find it easier to handle various calculus problems that involve finding slopes and rates of change.
Tangent Line
A tangent line to a curve at a specific point is a straight line that touches the curve just at that point and has the same slope as the curve at that point. It's like a snapshot of the curve's direction at that exact moment. To find the equation of a tangent line, you need two things: a point on the curve and the slope of the curve at that point.
In this exercise, we have the point \((2, 2)\) which lies on the curve \(y = 3x^2 - 5x\). We also know the slope from the derivative evaluation: \(f'(2) = 7\). Therefore, the tangent line has a slope of 7 at point \((2, 2)\).
We use the point-slope form of a linear equation to write the equation for the tangent line: \(y - y_1 = m(x - x_1)\). Substituting the point \((2, 2)\) and the slope \(m = 7\), we get \(y - 2 = 7(x - 2)\). Simplifying, this becomes \(y = 7x - 12\). This line represents the tangent to the curve at the point \((2, 2)\).
In this exercise, we have the point \((2, 2)\) which lies on the curve \(y = 3x^2 - 5x\). We also know the slope from the derivative evaluation: \(f'(2) = 7\). Therefore, the tangent line has a slope of 7 at point \((2, 2)\).
We use the point-slope form of a linear equation to write the equation for the tangent line: \(y - y_1 = m(x - x_1)\). Substituting the point \((2, 2)\) and the slope \(m = 7\), we get \(y - 2 = 7(x - 2)\). Simplifying, this becomes \(y = 7x - 12\). This line represents the tangent to the curve at the point \((2, 2)\).
Derivative Evaluation
Derivative evaluation involves finding the rate at which a function changes at a specific point. This is essentially finding the slope of the tangent line to the curve at that point. In our example, the derivative function we found was \(f'(x) = 6x - 5\). By substituting \(x = 2\), we get \(f'(2) = 6(2) - 5 = 7\).
This result tells us that at the point \((2, 2)\) on the curve \(y = 3x^2 - 5x\), the slope of the tangent line is 7. This slope indicates how steep the tangent is. A positive slope means the function is increasing at that point, while a negative slope would indicate a decrease.
Evaluating a derivative at a particular point gives insight into the behavior of the function at that spot. It’s crucial for understanding how the function changes and for practical applications like tangents and normals to curves.
This result tells us that at the point \((2, 2)\) on the curve \(y = 3x^2 - 5x\), the slope of the tangent line is 7. This slope indicates how steep the tangent is. A positive slope means the function is increasing at that point, while a negative slope would indicate a decrease.
Evaluating a derivative at a particular point gives insight into the behavior of the function at that spot. It’s crucial for understanding how the function changes and for practical applications like tangents and normals to curves.
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