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In calculus, the first derivative of a function helps us understand how that function behaves over its domain. It tells us about the rate of change of the function with respect to its variable, in this case, 'x'. For the function given, \[ j(x)=5e^{-x} + \ln x, \]we apply the rules of derivatives to find the first derivative, \( j'(x) \). Let's break it down:

• The derivative of an exponential function \( 5e^{-x} \) is \( -5e^{-x} \). The negative sign comes from the chain rule applied to \( e^{-x} \).
• The derivative of a natural logarithm function \( \ln x \) is \( \frac{1}{x} \).

By combining these results, the first derivative of the function is:\[ j'(x) = -5e^{-x} + \frac{1}{x} \]This derivative function \( j'(x) \) will be crucial for us to discern the slope of the curve at any point 'x' within the domain (\( x > 0 \)). A positive slope indicates that the function is increasing, while a negative slope shows that the function is decreasing at that point.

The second derivative of a function gives us insight into the curvature of the graph. It determines whether the function is concave up or concave down in various sections. A positive second derivative means the graph is concave up, while a negative value means it's concave down. For our function:\[ j'(x) = -5e^{-x} + \frac{1}{x}, \]we take the derivative again to get the second derivative \( j''(x) \).

• First, differentiate \( -5e^{-x} \) to get \( 5e^{-x} \).
• Then, differentiate \( \frac{1}{x} \) to attain \( -\frac{1}{x^2} \).

Putting these together, we have:\[ j''(x) = 5e^{-x} - \frac{1}{x^2} \]The second derivative \( j''(x) \) helps in identifying intervals of concavity. Since understanding where the graph of a function curves is essential for analyzing its behavior, \( j''(x) \) plays a crucial role in our study of inflection points.

Inflection points signify where the graph of a function changes its curvature from concave up to concave down or vice versa. These points can be found where the second derivative changes sign. This involves setting the second derivative to zero or finding values where it is undefined.In our example, we have:\[ j''(x) = 5e^{-x} - \frac{1}{x^2} = 0 \]Solving this gives:\[ 5x^2 = e^x \]This equation is transcendental, meaning it is not easily solved analytically but requires numerical methods to nail down specific solutions for \( x \). Identifying the right \( x \) values where this holds true will point us to the potential inflection points.Inflection points are crucial as they offer key insight into the behavior of the graph. They can dictate major shifts in trends, indicating when and how the function's rate of change is accelerating or decelerating in different sections of its graph.