/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 95 Suppose \(f\) is differentiable ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose \(f\) is differentiable on an interval containing \(a\) and \(b,\) and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f .\) Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming the tangent lines are not parallel (see figure). (Figure cant copy) a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b,\) for real numbers \(a\) and \(b.\) b. If \(f(x)=\sqrt{x},\) show that \(c=\sqrt{a b},\) the geometric mean of \(a\) and \(b,\) for \(a>0\) and \(b>0.\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a>0\) and \(b>0.\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.

Short Answer

Expert verified
Question: Find the x-coordinate 'c' of the intersection point of the tangent lines at points P(a, f(a)) and Q(b, f(b)) on the graph of the given functions. a. f(x) = x² b. f(x) = √x c. f(x) = 1/x d. General differentiable function f Answer: a. c = (a + b) / 2 b. c = √(ab) c. c = (2ab) / (a + b) d. c = (f'(b)a - f'(a)b + f(b) - f(a)) / (f'(b) - f'(a))

Step by step solution

01

Find the derivatives of f(x) = x² at points P(a, f(a)) and Q(b, f(b))

Using the power rule, the derivative of f(x) = x² is f'(x) = 2x. The slopes of the tangent lines at points P and Q are: f'(a) = 2a and f'(b) = 2b.
02

Calculate the equation of the tangent lines at P and Q

Using the point-slope form of a line, we get the equation of the tangent lines at P and Q as: y - f(a) = f'(a)(x - a) and y - f(b) = f'(b)(x - b) y - a² = 2a(x - a) and y - b² = 2b(x - b)
03

Find the intersection point (c,d) of the tangent lines

To find the intersection point, we need to solve the system of linear equations from step 2 for x and y: 2a(x - a) + a² = 2b(x - b) + b² Expanding and simplifying, we get x = (a + b)/2. Therefore, c = (a + b)/2. For function f(x)=√x:
04

Find the derivatives of f(x) = √x at points P(a, f(a)) and Q(b, f(b))

Using the power rule, the derivative of f(x) = √x is f'(x) = 1/(2√x). The slopes of the tangent lines at points P and Q are: f'(a) = 1/(2√a) and f'(b) = 1/(2√b).
05

Calculate the equation of the tangent lines at P and Q

Using the same procedure as in the previous part, we get the equation of the tangent lines at P and Q as: y - √a = (1/(2√a))(x - a) and y - √b = (1/(2√b))(x - b)
06

Find the intersection point of the tangent lines

Solving the system of linear equations from step 2 for x and y, we get: (1/(2√a))(x - a) + √a = (1/(2√b))(x - b) + √b Expanding and simplifying, we get x = √(ab). Therefore, c = √(ab). For function f(x)=1/x:
07

Find the derivatives of f(x) = 1/x at points P(a, f(a)) and Q(b, f(b))

Using the power rule, the derivative of f(x) = 1/x is f'(x) = -1/x². The slopes of the tangent lines at points P and Q are: f'(a) = -1/a² and f'(b) = -1/b².
08

Calculate the equation of the tangent lines at P and Q

Using the same procedure as in the previous parts, we get the equation of the tangent lines at P and Q as: y - 1/a = (-1/a²)(x - a) and y - 1/b = (-1/b²)(x - b)
09

Find the intersection point of the tangent lines

Solving the system of linear equations from step 2 for x and y, we get: (-1/a²)(x - a) + 1/a = (-1/b²)(x - b) + 1/b Expanding and simplifying, we get x = (2ab) / (a + b). Therefore, c = (2ab) / (a + b). For a general differentiable function f:
10

Find the derivatives f'(a) and f'(b) at points P(a, f(a)) and Q(b, f(b))

We assume f'(a) and f'(b) are the slopes of the tangent lines at points P and Q.
11

Calculate the equation of the tangent lines at P and Q

Using the point-slope form of a line, we get the equation of the tangent lines at P and Q as: y - f(a) = f'(a)(x - a) and y - f(b) = f'(b)(x - b)
12

Find the intersection point of the tangent lines

Solving the system of linear equations from step 2 for x and y: f'(a)(x - a) + f(a) = f'(b)(x - b) + f(b) Rearranging and isolating 'x', we get: x = (f'(b)a - f'(a)b + f(b) - f(a)) / (f'(b) - f'(a)) Therefore, c = (f'(b)a - f'(a)b + f(b) - f(a)) / (f'(b) - f'(a)).

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Key Concepts

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

Differentiable Functions
Understanding differentiable functions is integral when studying the behavior of curves, especially in calculus. A function is said to be differentiable at a point if it has a derivative there, which means that there is a well-defined tangent to the curve at that point. Graphically, differentiability ensures that the curve has no sharp corners or breaks at the point in question.

For a function to be differentiable over an interval, it must be differentiable at every point within that interval. This means that the function is smooth and continuous, allowing for the computation of slopes at every point. In the context of the exercise provided, for functions like f(x) = x^2, f(x) = √x, and f(x) = 1/x, the differentiability enables the calculation of tangent lines and their point of intersection which is crucial for such problems. Differentiability also implies continuity, but the reverse is not always true; a function may be continuous but not differentiable at certain points. For instance, the absolute value function is continuous everywhere but not differentiable at zero.
Mean Value Theorem
The Mean Value Theorem (MVT) is a fundamental theorem in differential calculus that provides a connection between derivatives and differences of values of a function. It essentially states that for any differentiable function that is continuous on a closed interval [a, b], there exists at least one point c in the open interval (a, b) such that the derivative at c is equal to the average rate of change of the function over the interval from a to b.

The formula given by MVT is f'(c) = (f(b) - f(a)) / (b - a). This theorem is akin to saying that there is at least one point at which the instantaneous rate of change (slope of the tangent) matches the average rate of change (slope of the secant joining (a, f(a)) and (b, f(b))). In the exercise, the MVT could be used to verify the existence of a tangent line with a specific slope that validates the conditions of the tangent intersection points for different functions.
Derivatives of Functions
Derivatives represent the rate at which a function is changing at any given point and are fundamental in calculus for studying all kinds of changes. The derivative of a function at a point is visually represented by the slope of the tangent line to the function's graph at that point. Derivatives can be computed using various rules, such as the power rule, product rule, quotient rule, and chain rule, depending on the complexity of the function.

In the context of the exercise, the power rule is used for finding the derivatives of functions like f(x) = x^2 and f(x) = 1/x. This rule states that if f(x) = x^n, then f'(x) = nx^(n−1). For example, the derivative of f(x) = x^2 is f'(x) = 2x. Similarly, derivatives help us find the equations of the tangent lines at points P and Q and subsequently allow us to determine the intersection point of these tangents, leading us to the value of c in the various scenarios presented in the exercise.

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