Question

Find the function The following limits represent the slope of a curve \(y=f(x)\) at the point \((a, f(a)) .\) Determine a possible function \(f\) and number \(a ;\) then calculate the limit. $$\lim _{x \rightarrow 1} \frac{3 x^{2}+4 x-7}{x-1}$$

Short answer

Answer: The function f(x) is f(x) = 3x^2 + 4x - 7, and the number a is 1.

Step-by-step solution

Identify the function f(x) and the number a

From the given limit: $$\lim _{x \rightarrow 1} \frac{3 x^{2}+4 x-7}{x-1}$$ We notice that the denominator \((x-1)\) represents the \(x-a\) part of the definition. Therefore, we can conclude that \(a = 1\). Now, by comparing the numerator of the given limit to the numerator of the definition, we see that: $$3x^2 + 4x - 7 = f(x) - f(1)$$ To solve for \(f(x)\), we need to find an antiderivative of the expression on the left, and then find \(f(1)\). We can refer the polynomial as \(g(x)\), such as \(g(x) = 3x^2 + 4x - 7\), then find \(g'(x)\) which is the derivative of \(g(x)\).

Find the derivative of g(x)

Since we know \(g(x) = 3x^2 + 4x - 7\), we can find its derivative by differentiating each term in the polynomial. $$g'(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(4x) - \frac{d}{dx}(7) = 6x + 4$$ Now, we have: $$g'(x) = 6x + 4$$

Find a possible function f(x)

Now find \(g(1)\): $$g(1) = 3(1)^2 + 4(1) - 7 = 3 + 4 - 7 = 0$$ Now, we know that \(g(x) = f(x) - f(1)\), so it becomes clear that \(f(1) = g(1) = 0\). Therefore, we can conclude that: $$f(x) = g(x) = 3x^2 + 4x - 7$$

Calculate the limit

Now that we have found the function \(f(x) = 3x^2 + 4x - 7\) and the number \(a = 1\), we can calculate the limit by substituting these values into the derivative formula: $$\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1} = \lim _{x \rightarrow 1} \frac{3x^2 + 4x - 7}{x-1}$$ Since we've already found the derivative, it would be simple to evaluate the limit at \(x=1\): $$g'(1) = 6(1) + 4 = 10$$ Therefore, the limit is: $$\lim _{x \rightarrow 1} \frac{3 x^{2}+4 x-7}{x-1} = 10$$

Key concepts

Derivative of Polynomial

Understanding the derivative of a polynomial is fundamental to calculus. Derivatives measure how a function changes as its input changes. For polynomials, which are functions that consist of variables raised to whole number powers and multiplied by coefficients, finding the derivative is a matter of applying some straightforward rules.

Let's consider a generic polynomial function such as \( P(x) = a_nx^n + a_{n-1}x^{n-1} + \.\.\. + a_1x + a_0 \). To find the derivative, denoted by \( P'(x) \), you apply the power rule: multiply the coefficient by the exponent and decrease the exponent by one. This process is repeated for each term in the polynomial. So, the derivative of \( P(x) \) would be \( P'(x) = na_nx^{n-1} + (n-1)a_{n-1}x^{n-2} + \.\.\. + a_1 \).

In the context of our exercise, the function \( g(x) = 3x^2 + 4x - 7 \) is a polynomial where the derivative, \( g'(x) \), is found using the power rule. Differentiating each term separately yields \( g'(x) = 6x + 4 \). In practical terms, now having \( g'(x) \) allows us to discuss the next crucial concept - the slope of a curve at a particular point.

Slope of a Curve

The slope of a curve at a particular point is a measure of how steep the curve is rising or falling at that point. In a graphical sense, it's the inclination of the tangent line to the curve at that point. For a function \( y = f(x) \), the slope at a point \( x = a \) is given by the derivative of the function evaluated at that point, denoted as \( f'(a) \).

The process of finding this slope is what the limit equation in our exercise aims to showcase. For the function \( f(x) \), the slope at the point \( a \) can be found by calculating the expression \( \lim _{x \rightarrow a} \frac{f(x)-f(a)}{x-a} \). This limit represents the derivative of \( f(x) \) at \( x = a \) or simply the slope of the curve at that point.

Returning to our example where \( f(x) = 3x^2 + 4x - 7 \), by finding that \( g'(1) = 10 \), we've actually determined the slope of the curve at the point where \( x = 1 \). This tells us how fast the function is increasing at the point \( x = 1 \) and provides the gradient of the tangent line to the curve at that specific point.

Rate of Change

The rate of change in a function is a concept that captures the idea of how quickly the value of the function changes as the input changes. It's a fundamental idea in calculus, closely related to derivatives and slopes of curves.

In its simplest form, the average rate of change of a function over an interval \( [a, b] \) is computed as \( \frac{f(b)-f(a)}{b-a} \), which resembles the slope formula for a straight line. When the points come infinitely close, the average rate of change becomes the instantaneous rate of change, which is the derivative of the function at a point.

In our previous discussion of the slope of a curve, we concluded that the derivative at a point represents the slope of the tangent line at that point. This is also the instantaneous rate of change of the function at that specific input value. In other words, the derivative not only gives us the steepness of the curve but also tells us how fast the function's output is changing at each point along the curve. For the polynomial function in our exercise, the expression \( g'(1) = 10 \) not only represented the slope but also indicated the rate at which \( f(x) \) was changing when \( x \) was at 1.