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Chapter 3: Problem 26

Given the function \(f\) and the point \(Q,\) find all points \(P\) on the graph of \(f\) such that the line tangent to \(f\) at \(P\) passes though \(Q\). Check your work by graphing \(f\) and the tangent lines. $$f(x)=-x^{2}+4 x-3 ; Q(0,6)$$

Short Answer

Expert verified
Answer: The coordinates of points P are (3, -2) and (-1, -6).

Step by step solution

01

Find the derivative of the function

To find the derivative of the function \(f(x) = -x^2 + 4x - 3\), use the power rule, which states that the derivative of \(x^n\) is \(nx^{n-1}\). $$f'(x) = \frac{d}{dx}(-x^2 + 4x - 3) = -2x + 4$$
02

Determine the equation of the tangent line

Using the point-slope form of a line (\(y-y_1=m(x-x_1)\)), we can write the equation of the tangent line at point P using the slope m (from the derivative \(f'(x)\)) and the point Q(0,6): $$y-6 = f'(x)(x-0) \Rightarrow y = -2x + 4x$$
03

Set up and solve the equation

Now, equate the function \(f(x)\) and the tangent line to set up an equation: $$-x^2 + 4x - 3 = -2x + 4x$$ This equation simplifies to: $$x^2 - 2x - 3 = 0$$ To solve for x, we can factor the quadratic: $$(x-3)(x+1) = 0$$ So, the two possible values of \(x\) are \(x=3\) and \(x=-1\). To find the corresponding y-values, plug these into the function \(f(x)\): $$f(3) = -(3)^2 + 4(3) - 3 = -2$$ $$f(-1) = -(-1)^2 + 4(-1) - 3 = -6$$ So the two points P are \((3, -2)\) and \((-1, -6)\).
04

Verify the solution by graphing

To confirm that the tangent lines pass through point Q, graph the function \(f(x)\) along with the tangent lines at points P. Observe that both tangent lines pass through the given point Q(0,6), which verifies the correctness of our solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative of a Function
The derivative of a function represents the rate at which the function's value is changing at a certain point. In calculus, this concept is fundamental because it gives us an immediate rate of change, much like speed is the immediate rate of change of distance with respect to time. When we look at the function f(x) = -x^2 + 4x - 3, we determine the derivative by applying the power rule.

The power rule tells us to multiply the exponent by the coefficient and reduce the exponent by one. Applying this to each term in the function f(x) gives us: f'(x) = -2x + 4. This derivative f'(x) now represents the slope of the tangent line to the graph of f at any point x.
Point-Slope Form of a Line
The point-slope form of a line is a straightforward method for writing the equation of a line when you know a point it passes through and its slope. The form is written as y - y1 = m(x - x1), where (x1, y1) is the point on the line, and m is the slope of the line.

In our problem, we use the point-slope form to write the equation of the tangent line at any point P on the curve by taking point Q and the slope given by the derivative at P. The point Q(0, 6) and the derivative f'(x) provide all the information we need to construct the tangent line's equation.
Quadratic Equation Solving
Solving a quadratic equation is a fundamental skill in algebra, typically done by factoring, completing the square, or using the quadratic formula. In our calculus problem, the equation arises from setting the expression for the function equal to the expression for the derivative, simplifying to a standard quadratic form: ax^2 + bx + c = 0.

Factoring is usually the quickest method when possible. In this case, our equation x^2 - 2x - 3 = 0 factors neatly into (x - 3)(x + 1) = 0. Once factored, we set each factor equal to zero to solve for x, resulting in the possible solutions for points where the tangent passes through Q.
Graphing Functions and Tangents
Graphing functions and their tangent lines helps us visualize the behavior of those functions and the concept of the derivative. After calculus procedures, such as finding the derivative and solving equations, graphing serves as a confirmation tool.

In the context of our problem, we graph the quadratic function f(x) = -x^2 + 4x - 3 and then plot the tangent lines at the points of tangency we've calculated. Seeing that these tangent lines indeed pass through the point Q(0, 6) verifies our work. Graphing also helps students intuitively understand the relationship between a function and its derivative.

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Most popular questions from this chapter

Identifying functions from an equation. The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots\) c. Use the functions found in part (b) to graph the given equation. \(y^{3}=a x^{2}(\text { Neile's semicubical parabola })\)

Prove the following identities and give the values of \(x\) for which they are true. $$\cos \left(2 \sin ^{-1} x\right)=1-2 x^{2}$$

Tangent lines and exponentials. Assume \(b\) is given with \(b>0\) and \(b \neq 1 .\) Find the \(y\) -coordinate of the point on the curve \(y=b^{x}\) at which the tangent line passes through the origin. (Source: The College Mathematics Journal, 28, Mar 1997).

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x^{2}(y-2)-e^{y}=0$$

A store manager estimates that the demand for an energy drink decreases with increasing price according to the function \(d(p)=\frac{100}{p^{2}+1},\) which means that at price \(p\) (in dollars), \(d(p)\) units can be sold. The revenue generated at price \(p\) is \(R(p)=p \cdot d(p)\) (price multiplied by number of units). a. Find and graph the revenue function. b. Find and graph the marginal revenue \(R^{\prime}(p)\). c. From the graphs of \(R\) and \(R^{\prime}\), estimate the price that should be charged to maximize the revenue.
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