Problem 3
Question
Let \(f(x)=x^{3}\) and \(P\) be the point \((1,1)\) on the graph of \(f\). (a) Approximate the slope of the line tangent to \(f\) at \(P\) by looking at the slope of the secant line through \(P\) and \(Q\), where \(Q=(1+h, f(1+h)\) ). Calculate the difference quotient for various values of \(h\), both positive and negative. See if your calculator or computer will produce a table of values. (b) Calculate \(f^{\prime}(1)\) by computing the limit of the difference quotient. (c) For what values of \(h\) is the difference quotient greater than \(f^{\prime}(1) ?\) For what values of \(h\) is the difference quotient less than \(f^{\prime}(1) ?\) Make sense out of this by looking at the graph of \(x^{3}\).
Step-by-Step Solution
Verified Answer
The slope of the tangent line to the function \(f(x) = x^3\) at point (1,1) is 3. The difference quotient will be slightly more than this slope when \(h\) is a small positive number, and slightly less when \(h\) is a small negative number.
1Step 1: Calculate the difference quotient for various values of h
The difference quotient for the function \(f(x) = x^{3}\) will be \(\frac{f(1+h)-f(1)}{h}\) which simplifies to \(\frac{(1+h)^3-1^3}{h}\). Now, we can plug in various values for \(h\) to find the approximate slope of secant line through \(P\) and \(Q\). The important thing here is choosing very small positive and negative values for \(h\), as they give an approximation of the derivative which is the slope of the tangent line.
2Step 2: Compute \(f^{\prime}(1)\)
The derivative of \(f(x) = x^3\) is \(f^{\prime}(x) = 3x^2\). If we plug in \(x = 1\), we get \(f^{\prime}(1) = 3\). This is the slope of the tangent line at the point (1,1).
3Step 3: Comparison of difference quotient with \(f^{\prime}(x)\)
As you vary \(h\), you will witness that for small positive \(h\), the difference quotient is slightly more than \(f^{\prime}(1)\) and for small negative \(h\), it is less than \(f^{\prime}(1)\). This is sensible as if you move a tiny bit away from the point, the slope of secant line (difference quotient) increases a bit as the graph curves upwards, and it decreases if you move to the left of the given point for the same reason.
Key Concepts
Tangent Line SlopeSecant Line to Tangent Line ApproximationCalculus Limit Definition of Derivative
Tangent Line Slope
Understanding the slope of a tangent line is crucial when exploring the rates at which functions change. It represents the instantaneous rate of change of a function at a certain point. Imagine you're zooming in as close as you can on a curve at a point until the curve looks straight. The slope of this 'flattened' line is the slope of the tangent.
For example, if we look at the function
As we compute the difference quotient for smaller and smaller values of
For example, if we look at the function
f(x) = x^3 at point P(1,1), we need to find the slope of the line that just grazes the curve at this point. Visually, this line would just touch the curve without crossing it. One robust method to calculate this slope is by using the difference quotient for values of h very close to zero which gives us an approximation of the slope.As we compute the difference quotient for smaller and smaller values of
h and approach zero, our secant lines get closer to actually being the tangent line, and the slope of these secant lines approaches the true tangent slope. This is how calculus bridges the gap between secants and tangents, offering a precise tool to measure instantaneous change.Secant Line to Tangent Line Approximation
The concept of transitioning from a secant line to a tangent line involves understanding how a line crossing two points on a curve can become the line that just touches it at a single point when those two points are infinitely close. The difference quotient, which is
With
(f(a+h)-f(a))/h, provides a formula for the slope of the secant line.With
f(x) = x^3 and a point P(1,1), the secant line between P and another point Q(1+h, (1+h)^3) gives an approximation of the tangent slope as h approaches zero. Initially, this quotient for various small values of h will not be exactly three, but as we choose smaller values for h, positive and negative, the quotient will get closer to the derivative's value at x=1, which is the tangent's slope at point P. This serves as a practical estimate for the actual slope of the tangent line.Calculus Limit Definition of Derivative
The limit definition of the derivative stands at the heart of calculus, paving the way from the average rate of change to the instantaneous. It elegantly summarizes the process of finding tangent slopes using the limit of the difference quotient as
For our function,
This process solidifies our understanding of derivatives not just as a formula, but as a fundamental principle of how functions change, and it's a concept illuminated by the limit of the difference quotient as
h approaches zero.For our function,
f'(x) = 3x^2, when we compute f'(1), we're essentially doing the following: We're taking the limit as h goes to zero of the expression (f(1+h) - f(1))/h. This expression is the average rate of change over the interval from x to x + h, and as h gets infinitesimally small, this average rate turns into the instantaneous rate — the derivative, which in this case equates to three. This process solidifies our understanding of derivatives not just as a formula, but as a fundamental principle of how functions change, and it's a concept illuminated by the limit of the difference quotient as
h approaches zero.Other exercises in this chapter
Problem 2
Let \(f(x)=\frac{1}{x} .\) In this problem we will look at the slope of the tangent line to \(f(x)\) at point \(P=\left(\frac{1}{2}, 2\right)\) (a) Is the slope
View solution Problem 2
Use the limit de nition of derivative to nd the derivative of \(f(x)=k x^{2}\).
View solution Problem 3
Let \(f(x)=x^{2}\). Find the point at which the line tangent to \(f(x)\) at \(x=2\) intersects the line tangent to \(f(x)\) at \(x=-1\).
View solution Problem 4
Let \(f(x)=x^{3}\) and \(P\) be the point \((0,0)\) on the graph of \(f\). (a) Approximate \(f^{\prime}(0)\) by looking at the slope of the secant line through
View solution