Problem 9 - Single Variable Calculus

Question

(a) use Equation (1) to find the slope of the secant line passing through the points \((a, f(a))\) and \((a+h, f(a+h)) ;\) (b) use the results of part (a) and Equafion (2) to find the slope of the tangent line at the point \((a, f(a)) ;\) and \((\mathrm{c})\) find an equation of the tangent line to the graph of \(f\) at the point \((a, f(a))\). \(f(x)=2 x^{2}-1 \quad(2,7)\)

Step-by-Step Solution

Answer

\(y - 7 = \frac{8}{-2+\sqrt{14}} (x - 2)\)

Step 1: Find the value of h

We are given the point \((2, 7)\). Since the point lies on the curve, that means for \(x=a=2\), \(f(a)=7\). Now, we substitute this value in the equation \(f(x)=2x^2-1\), i.e., \(7 = 2(2)^2 - 1\), and solve for \(h\): \[7=2(2+h)^2-1\] Now, let's solve for \(h\).

Step 2: Solve for h

To solve for \(h\), firstly expand the equation: \[7=2(4+4h+h^2)-1\] Now, simplify the equation: \[7=8+8h+2h^2-1\] \[0=2h^2+8h+1\] Now, we have a quadratic equation. We can solve for \(h\) using the quadratic formula, where \(a = 2\), \(b = 8\), and \(c = 1\): \[h=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] \[h=\frac{-8\pm \sqrt{8^2-4\cdot 2 \cdot 1}}{4}\] \[h=\frac{-8\pm \sqrt{64-8}}{4}\] \[h=\frac{-8\pm \sqrt{56}}{4}\] \[h=\frac{-2\pm \sqrt{14}}{1}\] As we have two values for \(h\), we need to choose only one. The one that leads to a smaller secant line is usually chosen. Since the negative value of \(h\) \((\)i.e., \(h = -2 + \sqrt{14})\) will lead to smaller secant lines, we will proceed with this value

Step 3: Find the slope of the secant line

Using Equation (1), we will find the slope of the secant line, \(m_{sec}\), using the points \((a, f(a))\) and \((a+h, f(a+h))\): \[m_{sec}=\frac{f(a+h)-f(a)}{h}\] Using the given function \(f(x) = 2x^2 - 1\), we will substitute the values of \(a\), \(f(a)\), and \(h\) into the equation: \[m_{sec}=\frac{2(a+h)^2-1-(2a^2-1)}{h} \] \[m_{sec}=\frac{2(2-h)^2-1-(2(2)^2-1)}{-2+\sqrt{14}}\] Now, simplify the expression: \[m_{sec}=\frac{8(1-h)(1+h)}{-2+\sqrt{14}}\] Thus, we found the slope of the secant line, \(m_{sec}\).

Step 4: Find the slope of the tangent line

Now, let's use Equation (2) to find the slope of the tangent line, \(m_{tan}\). We can obtain Equation (2) by taking the limit of the slope of the secant line as \(h\) approaches to \(0\). \[m_{tan} = \lim_{h \to 0} m_{sec}\] \[m_{tan} = \lim_{h \to 0} \frac{8(1-h)(1+h)}{-2+\sqrt{14}}\] As \(h\) approaches \(0\), the expression becomes: \[m_{tan} = \frac{8(1)(1)}{-2+\sqrt{14}}\] Simplifying, we get: \[m_{tan} = \frac{8}{-2+\sqrt{14}}\] So, we found the slope of the tangent line, \(m_{tan}\).

Step 5: Find the equation of the tangent line

Finally, let's find the equation of the tangent line. We will use the point-slope form of a line, with the point \((a, f(a))\) and the slope \(m_{tan}\): \[y - f(a) = m_{tan}(x - a)\] Substituting the values for \(a\), \(f(a)\), and \(m_{tan}\), we get: \[y - 7 = \frac{8}{-2+\sqrt{14}} (x - 2)\] Now, you can leave the equation of the tangent line as it is or you can simplify it further by multiplying both sides by the denominator to get rid of the fraction.

Key Concepts

Secant Line

A secant line is a straight line that intersects a curve at two or more points. In calculus, the secant line helps us approximate the slope of a curve over a specific interval.
For a given function, let's say we're looking at a curve described by the function \( f(x) \). We pick two points on this curve: \((a, f(a))\) and \((a+h, f(a+h))\). The slope of the secant line connecting these two points can be calculated using the formula:

• \( m_{sec} = \frac{f(a+h) - f(a)}{h} \)

This formula gives us the average rate of change of the function over the interval from \( a \) to \( a+h \), which is essentially the slope of the curve between these two points.
Secant lines are particularly useful because they can provide insight into how steep or flat a curve is over a specific span. The closer the points on the curve, the more accurately the secant line represents the curve's instantaneous rate of change at that interval.

Tangent Line

A tangent line is a straight line that touches a curve at exactly one point without crossing it at that location. Unlike a secant line, which intersects a curve at multiple points, a tangent line represents the instantaneous rate of change at a specific point.
In mathematical terms, the slope of a tangent line at a point \((a, f(a))\) on the curve \(f(x)\) is determined by the limit of the slope of the secant line as \(h\) approaches zero. We write this as:

• \( m_{tan} = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)

This is essentially the "derivative" of the function at the point \( a \). The tangent line can be used to approximate the function around that point. If the function is close to a straight line at that point, the approximation will be very accurate.
For example, if we have a point \((2,7)\) on the curve \(f(x) = 2x^2 - 1\), the tangent line at this point will give us a straight line that approximates the curve near \(x=2\).

Derivative

The derivative of a function is a concept that represents the rate of change of the function with respect to one of its variables. It gives us the slope of the tangent line at any point on the function's graph. The derivative is fundamental in calculus as it tells us how a function behaves as its input changes.
To find the derivative of a function \( f(x) \) at a point, you use the same formula as for the slope of a tangent line:

• \( f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \)

This limit gives the precise value of what the function's rate of change is when \(h\) is infinitely small, offering a more precise picture than a secant line.
Derivatives have countless applications in fields like physics, engineering, and economics, where understanding how quantities change is crucial.

Quadratic Formula

The quadratic formula is a method used to find the solutions of a quadratic equation, which takes the general form \( ax^2 + bx + c = 0 \). This formula is particularly useful when factoring a quadratic equation is challenging.
The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula computes the roots by solving for \(x\) in terms of the coefficients \(a\), \(b\), and \(c\).
Each solution (or "root") of the equation corresponds to a point where the graph of the quadratic function intersects the x-axis. In other words, these points are the values of \(x\) for which the function \( f(x) = ax^2 + bx + c \) equals zero.
Using the quadratic formula, any quadratic equation can be easily solved, which makes it a powerful tool not only in algebra but also in calculus, especially when dealing with curves described by quadratic functions.