Given The Function $f(x) = 6x + 7 - 5e^x$ And The Tangent Line To The Graph Of $f(x)$ At The Point $(0, 2)$ Expressed As $y = Mx + B$, Find The Values Of $m$ And $b$.

Jun 21, 2025 by ADMIN 167 views

Finding the Tangent Line to f(x) = 6x + 7 - 5e^x at (0, 2)

This article delves into the process of determining the equation of the tangent line to the function f(x) = 6x + 7 - 5e^x at the specific point (0, 2). We will explore the fundamental concepts of calculus, specifically derivatives, to calculate the slope of the tangent line. Then, we'll utilize the point-slope form of a linear equation to construct the tangent line's equation, expressing it in the familiar slope-intercept form, y = mx + b. This involves calculating the slope (m) and the y-intercept (b) to fully define the tangent line.

Determining the Slope (m)

The cornerstone of finding the tangent line lies in determining its slope, often denoted as m. In calculus, the derivative of a function at a specific point represents the slope of the tangent line at that point. Therefore, our initial task is to find the derivative of the function f(x) = 6x + 7 - 5e^x. To achieve this, we will employ the fundamental rules of differentiation.

The derivative of a sum or difference of terms is simply the sum or difference of the derivatives of each individual term. This allows us to break down the function into manageable parts. We will differentiate each term separately and then combine the results. The derivative of 6x is straightforward; it's simply 6, following the power rule of differentiation. The derivative of a constant, such as 7, is always 0. Lastly, we need to find the derivative of -5e^x. The derivative of e^x is e^x itself, and the constant multiple rule states that the derivative of a constant times a function is the constant times the derivative of the function. Therefore, the derivative of -5e^x is -5e^x.

Combining these individual derivatives, we obtain the derivative of the function f(x), denoted as f'(x):

f'(x) = 6 - 5e^x

Now that we have the derivative function, we can determine the slope of the tangent line at the point (0, 2). We do this by evaluating the derivative f'(x) at x = 0. This gives us the instantaneous rate of change of the function at that specific point, which is precisely the slope of the tangent line.

Substituting x = 0 into f'(x), we get:

f'(0) = 6 - 5e^0

Since e^0 = 1, the equation simplifies to:

f'(0) = 6 - 5(1) = 6 - 5 = 1

Therefore, the slope m of the tangent line at the point (0, 2) is 1. This value represents the steepness of the tangent line as it touches the curve of f(x) at that particular point.

Calculating the y-intercept (b)

Having determined the slope m of the tangent line, our next step is to calculate the y-intercept b. The y-intercept is the point where the line crosses the y-axis, and it's a crucial component in defining the line's position on the coordinate plane. To find b, we'll utilize the point-slope form of a linear equation and the information we already have: the slope m = 1 and the point (0, 2) where the tangent line touches the curve.

The point-slope form of a linear equation is given by:

y - y₁ = m(x - x₁)

where (x₁, y₁) is a point on the line and m is the slope. In our case, (x₁, y₁) is (0, 2) and m is 1. Substituting these values into the point-slope form, we get:

y - 2 = 1(x - 0)

Simplifying the equation, we have:

y - 2 = x

Now, to express the equation in the slope-intercept form (y = mx + b), we simply isolate y by adding 2 to both sides of the equation:

y = x + 2

Comparing this equation to the slope-intercept form, we can clearly see that the y-intercept b is 2. This confirms that the tangent line intersects the y-axis at the point (0, 2), which is consistent with the given point of tangency.

The Equation of the Tangent Line

With the slope m = 1 and the y-intercept b = 2 calculated, we can now definitively state the equation of the tangent line to the graph of f(x) = 6x + 7 - 5e^x at the point (0, 2). Using the slope-intercept form y = mx + b, we substitute the values of m and b to obtain:

y = 1x + 2

Which can be simplified to:

y = x + 2

This equation represents the unique line that touches the curve of f(x) at the point (0, 2) and has the same instantaneous rate of change as the function at that point. The tangent line provides a linear approximation of the function's behavior in the immediate vicinity of the point of tangency.

Summary of Results

In summary, we have successfully determined the equation of the tangent line to the function f(x) = 6x + 7 - 5e^x at the point (0, 2). Our calculations have yielded the following results:

Slope (m): 1
y-intercept (b): 2

Therefore, the equation of the tangent line is:

y = x + 2

This detailed process highlights the application of derivatives in finding tangent lines, a fundamental concept in calculus with wide-ranging applications in various fields.

Conclusion

In conclusion, finding the tangent line to a curve at a given point is a quintessential problem in calculus, illustrating the power and utility of derivatives. By meticulously calculating the derivative of the function f(x) = 6x + 7 - 5e^x and evaluating it at the point x = 0, we successfully determined the slope of the tangent line. Subsequently, by employing the point-slope form of a linear equation and the given point (0, 2), we calculated the y-intercept. This process culminated in the determination of the tangent line's equation, y = x + 2. This exercise not only reinforces the fundamental concepts of calculus but also showcases how these concepts can be applied to solve concrete problems.

The tangent line, as a linear approximation of the function at a specific point, serves as a valuable tool in various applications, including optimization problems, numerical analysis, and physics. Understanding how to find tangent lines is therefore a crucial skill for anyone delving into the realms of calculus and its applications. The steps outlined in this article provide a clear and concise methodology for tackling such problems, emphasizing the importance of a solid grasp of differentiation and linear equations. This problem serves as a microcosm of the broader applications of calculus in modeling and understanding the behavior of functions and their rates of change.