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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. $$f(x)=x^{2}+1 ; Q(3,6)$$

Short Answer

Expert verified
Answer: No, there are no points on the graph of the function \(f(x) = x^2 + 1\) such that the tangent line at those points passes through the point \(Q(3, 6)\).

Step by step solution

01

Find the derivative of the function

Find the derivative of the function \(f(x) = x^2 + 1\) with respect to x. This will give us the slope of the tangent line at any point \((x, f(x))\) on the graph of the function. We use the power rule to find the derivative: $$f'(x) = \frac{d}{dx}(x^2 + 1) = 2x$$
02

Set up the equation for the tangent line

Using the point-slope form of a line, we can write the equation for the tangent line at any point \((x, f(x))\) on the graph as follows: $$y - f(x) = f'(x)(x - 3)$$ Substitute \(f(x) = x^2 + 1\) and \(f'(x) = 2x\) into the equation: $$y - (x^2+1) = 2x(x - 3)$$ We know that this tangent line passes through the point \(Q(3, 6)\). Substitute the coordinates of \(Q\) into the equation: $$6 - (3^2+1) = 2(3)(3 - 3)$$ This simplifies to: $$6 - 10 = 0$$ This is not possible, so there is no point on the graph where the tangent line passes through the point \(Q(3, 6)\).
03

Check your work with graphing

To check our answer, we can graph the function \(f(x) = x^2 + 1\) and confirm there is no point on the graph where the tangent line passes through \(Q(3, 6)\). Using a graphing calculator or software, graph the function and observe that there is no tangent line at any point on the curve that passes through \(Q(3, 6)\), which agrees with our conclusion from the previous step. There are no points on the graph of the function \(f(x) = x^2 + 1\) such that the tangent line at those points passes through the point \(Q(3, 6)\).

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Key Concepts

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

Derivative
In calculus, the derivative of a function represents the rate at which the function's value changes with respect to changes in the input value. It essentially provides the slope of the function at any given point. In mathematical terms, if you have a function \( f(x) \), its derivative is denoted as \( f'(x) \). This concept is crucial when analyzing how a function's output behaves and changes.

For the function \( f(x) = x^2 + 1 \), we calculate the derivative to find the slope at any point on the curve. Applying the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \), we find:

\[ f'(x) = \frac{d}{dx}(x^2 + 1) = 2x \]

This result tells us that the slope of the tangent line to the function at point \( x \) is \( 2x \). Understanding the derivative allows us to effectively analyze the behavior of functions graphically and practically.
Tangent Line
A tangent line is a straight line that touches a function's graph at exactly one point and has the same slope as the function at that point. This unique property ensures that the tangent line provides a local linear approximation of the function around the point of tangency. Essentially, the tangent line is the best straight-line approximation of a curve at a particular point.

To find the equation of a tangent line at a point \( (x, f(x)) \) on the curve, we can use the point-slope form:
\[ y - f(x) = f'(x)(x - x_0) \]
where \( f'(x) \) is the slope at that point, as obtained from the derivative, and \( (x_0, y_0) \) is the external point through which this tangent line is required to pass.

In our exercise, we attempted to find if any tangent line of \( f(x) = x^2 + 1\) passes through the point \( Q(3, 6) \). Despite setting up the equation, there was no valid solution indicating that no such tangent line exists.
Function Graph
The graph of a function provides a visual representation of the relationship between the input values \(x\) and the output values \(f(x)\). Each point on the graph corresponds to a pair \((x, f(x))\) that satisfies the function equation. Graphical analysis is indispensable for understanding the behavior and properties of functions such as continuity, symmetry, and critical points.

For the function \( f(x) = x^2 + 1 \), the graph is a parabola that opens upwards with its vertex at \((0, 1)\). This shape indicates that as \( x \) increases or decreases from the vertex, \( f(x) \) increases rapidly.

Graphing the function alongside the potential tangent lines provides a concrete way to verify mathematical calculations. In this case, visual inspection would confirm the conclusion that there is no point on the curve from which a tangent line would pass through \( Q(3, 6) \). This reinforces the correctness of the derivative and tangent line calculations.

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