Problem 3 - Calculus and its applications

Question

a) Graph the function. b) Draw tangent lines to the graph at points whose \(x\) -coordinates are \(-2,0,\) and 1 c) Find \(f^{\prime}(x)\) by determining \(\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\). d) Find \(f^{\prime}(-2), f^{\prime}(0),\) and \(f^{\prime}(1) .\) These slopes should match those of the lines you drew in part (b). $$f(x)=-2 x^{2}$$

Step-by-Step Solution

Answer

Derivative: \(f'(x) = -4x\). Slopes are 8, 0, and -4 at \(x = -2, 0, 1\).

Step 1: Graphing the Function

To start, plot the function \( f(x) = -2x^2 \). This is a downward-opening parabola with its vertex at the origin (0,0). The parabola is vertically stretched by a factor of 2.

Step 2: Drawing Tangent Lines

Next, draw tangent lines on the graph at the points \( x = -2, x = 0, \) and \( x = 1 \). At each of these points, the tangent line must touch the parabola at only one point and have a slope equal to the derivative of \( f(x) \) at that point.

Step 3: Calculate the Derivative Symbolically

Use the definition of the derivative to find \( f'(x) \) by calculating the limit:\[f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{-2(x+h)^2 + 2x^2}{h}\]Expanding, we have:\[(x+h)^2 = x^2 + 2xh + h^2\]So the expression becomes:\[\lim_{h \to 0} \frac{-2(x^2 + 2xh + h^2) + 2x^2}{h}\]\[= \lim_{h \to 0} \frac{-2x^2 - 4xh - 2h^2 + 2x^2}{h}\]\[= \lim_{h \to 0} \frac{-4xh - 2h^2}{h}\]\[= \lim_{h \to 0} (-4x - 2h)\]As \( h \to 0 \), we find:\[f'(x) = -4x\]

Step 4: Evaluate Derivative at Specific Points

To find \( f'(-2) \), \( f'(0) \), and \( f'(1) \), substitute these \( x \)-values into the expression for the derivative:- \( f'(-2) = -4(-2) = 8 \)- \( f'(0) = -4(0) = 0 \)- \( f'(1) = -4(1) = -4 \)

Step 5: Match Slopes of Tangent Lines

Finally, verify that the slopes of the tangent lines drawn in Step 2 match the calculated derivative values:- At \( x = -2 \), slope of tangent: 8- At \( x = 0 \), slope of tangent: 0- At \( x = 1 \), slope of tangent: -4These match perfectly with the calculated \( f'(-2) \), \( f'(0) \), and \( f'(1) \).

Key Concepts

Graphing Functions

When you graph a function like \( f(x) = -2x^2 \), you're working with a quadratic function. This particular function represents a downward-opening parabola. Here's why: the coefficient of \( x^2 \), which is \(-2\), is negative, causing the parabola to open downwards.

The vertex of the parabola for any function of the form \( ax^2 + bx + c \) is at \( (0,0) \) when \( b \) and \( c \) are both zero. In our case, the parabola is centered at the origin. The "-2" affects the width of the parabola, making it narrower due to vertical stretching. The vertical stretch happens because any value of \( x \) is multiplied by a larger absolute number, here it is -2, which pulls the graph closer to the vertical axis.

Tangent Lines

Tangent lines are straight lines that touch a curve at just one point. When we draw tangent lines to the function \( f(x) = -2x^2 \) at \( x = -2, 0, \) and \( 1 \), these lines indicate the slope of the function at those specific points.

To sketch a tangent line, it should:

• Touch the curve at exactly one point.
• Have the same slope as the function at that point.

Understanding tangent lines helps in visualizing how the function behaves locally at a point, and it serves as a foothold for finding derivatives. The points where these tangent lines touch are called points of tangency, and the slope of each line visually represents the rate of change of the function at those points.

Limit Definition of Derivatives

The derivative of a function represents its rate of change or how much it changes as the input (\( x \)) changes. We define the derivative using limits, often referred to as the "limit definition of a derivative."

For the function \( f(x) = -2x^2 \), calculating the derivative \( f'(x) \) involves finding the limit:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]This starts by expanding \( (x+h)^2 \) and simplifying terms to isolate \( h \). The process involves:

• Expanding \( (x+h)^2 \) to get \( x^2 + 2xh + h^2 \).
• Substituting back and canceling terms to get the expression with \( h \) on the bottom.
• Taking the limit as \( h \to 0 \) to simplify further, leaving us with the derivative function.

This process shows how small changes (via the \( h \) approaching 0) affect the function, ultimately giving us \( f'(x) = -4x \).

Polynomial Functions

Polynomial functions, like \( f(x) = -2x^2 \), are algebraic expressions consisting of variables and coefficients. These functions can have multiple terms, each with variable powers expressed in integers.

Here's what you need to know about polynomial functions:

• The degree of the polynomial is the highest power of \( x \). In \( -2x^2 \), the degree is 2, making it a quadratic polynomial.
• A quadratic polynomial graphs as a parabola, which can open upwards or downwards depending on the leading coefficient's sign.
• They are continuous and smooth, meaning no sudden jumps or sharp corners, which makes them easier to graph and analyze.

Polynomial functions are fundamental in calculus because their derivatives are easy to compute using the power rule. They often serve as approximations for more complex functions, making them indispensable in the study of calculus.