Finding The Maximum Of A Logarithmic Function

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Hey guys! Let's dive into finding the maximum point of the function f(x)=log2(2+2xβˆ’x2)βˆ’2f(x) = log_2 (2+2x-x^2) - 2. This is a classic calculus problem, and we'll break it down step-by-step to make sure everything's crystal clear. We'll use our knowledge of logarithms and quadratic equations to solve this. Ready to get started? Let's go!

Understanding the Problem

So, our mission, should we choose to accept it, is to find the maximum value of the given function. But what does that even mean? Well, think of the graph of the function. We're trying to find the highest point on that curve. The function is f(x)=log2(2+2xβˆ’x2)βˆ’2f(x) = log_2 (2+2x-x^2) - 2. This looks a little tricky at first glance, because we see both a logarithm and a quadratic expression lurking inside. Understanding the components of the function is the first crucial step. The term log2(2+2xβˆ’x2)log_2 (2+2x-x^2) tells us that we're dealing with a logarithmic function. Remember, the logarithm is the inverse of the exponentiation. Specifically, this is a base-2 logarithm. The expression (2+2xβˆ’x2)(2+2x-x^2) is a quadratic equation, which means its graph is a parabola. The entire term is subtracting 2 which means we need to shift the whole graph down 2 units along the y axis. Therefore, the goal is to determine the x-value where f(x) reaches its highest value.

To make things easier to comprehend, let us separate the problem into smaller bits. It is easier to identify and understand the components individually. The quadratic term (2+2xβˆ’x2)(2+2x-x^2) will play a crucial role because the logarithm is only defined for positive values. So, first of all, we need to know the values of x for which the quadratic expression inside the logarithm, (2+2xβˆ’x2)(2+2x-x^2), is positive. This helps us to determine the domain of our function. And knowing the domain is super important because we can't look for a maximum outside of it!

Think about it this way: a logarithm is only defined for positive arguments. So, before we even start thinking about the maximum, we need to make sure we're working with valid x values. Finding the domain is the first thing that we need to do. Because the logarithm log2(z)log_2(z) is defined only when z>0z > 0, we need to find all x values for which 2+2xβˆ’x2>02 + 2x - x^2 > 0. We can rewrite this inequality as x2βˆ’2xβˆ’2<0x^2 - 2x - 2 < 0. Then by solving this quadratic equation for its roots, we'll find where the parabola crosses the x-axis. Using the quadratic formula, the roots are x=(2±√(4+8))/2=(2±√12)/2=1±√3x = (2 Β± √(4 + 8))/2 = (2 Β± √12)/2 = 1 Β± √3. Therefore the quadratic is less than 0 outside the interval (1βˆ’βˆš3,1+√3)(1 - √3, 1 + √3).

Finding the Domain of the Function

As we previously discussed, the domain is the set of all possible x values for which the function is defined. The logarithmic function, log2(u)log_2(u), is only defined for u>0u > 0. So, for our function f(x)=log2(2+2xβˆ’x2)βˆ’2f(x) = log_2 (2+2x-x^2) - 2, the expression inside the logarithm, 2+2xβˆ’x22+2x-x^2, must be greater than zero. That's our first constraint. How do we find that domain?

To find where 2+2xβˆ’x2>02+2x-x^2 > 0, we can rearrange the inequality as x2βˆ’2xβˆ’2<0x^2 - 2x - 2 < 0. Then we can use the quadratic formula to find the roots, i.e., the values of x for which x2βˆ’2xβˆ’2=0x^2 - 2x - 2 = 0. The quadratic formula gives us x=(βˆ’b±√(b2βˆ’4ac))/2ax = (-b Β± √(b^2 - 4ac)) / 2a. In our case, a=1a = 1, b=βˆ’2b = -2, and c=βˆ’2c = -2. So, plugging these values into the formula, we get:

x=(2±√((βˆ’2)2βˆ’4βˆ—1βˆ—βˆ’2))/(2βˆ—1)=(2±√(4+8))/2=(2±√12)/2=1±√3x = (2 Β± √((-2)^2 - 4 * 1 * -2)) / (2 * 1) = (2 Β± √(4 + 8)) / 2 = (2 Β± √12) / 2 = 1 Β± √3

So the roots are 1βˆ’βˆš31 - √3 and 1+√31 + √3. Because the quadratic expression x2βˆ’2xβˆ’2x^2 - 2x - 2 is a parabola opening upwards (because the coefficient of x2x^2 is positive), the quadratic expression is negative between the roots, which means x2βˆ’2xβˆ’2<0x^2 - 2x - 2 < 0 when 1βˆ’βˆš3<x<1+√31 - √3 < x < 1 + √3. Therefore, the domain of the function f(x)f(x) is the interval (1βˆ’βˆš3,1+√3)(1 - √3, 1 + √3).

It is essential to understand the domain of the function to avoid errors while calculating the maximum. When finding the maximum, we'll only consider values of x within this range. Remember, the domain is key! Make sure to calculate the domain before moving on, or it may impact the answers.

Finding the Critical Points

Now, let's look for critical points. Critical points are points where the derivative of the function is either zero or undefined. These are potential locations for maximum or minimum values. To find them, we need to calculate the derivative of f(x)f(x) with respect to x. Our function is f(x)=log2(2+2xβˆ’x2)βˆ’2f(x) = log_2 (2+2x-x^2) - 2.

First, we know that the derivative of loga(u)log_a(u) is uβ€²/(uβˆ—ln(a))u' / (u * ln(a)). Applying this rule to our function, we get:

fβ€²(x)=(2βˆ’2x)/((2+2xβˆ’x2)βˆ—ln(2))f'(x) = (2 - 2x) / ((2+2x-x^2) * ln(2))

Now we need to find where fβ€²(x)=0f'(x) = 0. This occurs when the numerator is zero. Therefore, we set 2βˆ’2x=02 - 2x = 0, which gives us x=1x = 1. This is a critical point. We must also check whether our critical point is within the domain or not. Because our domain is (1βˆ’βˆš3,1+√3)(1 - √3, 1 + √3), and 1βˆ’βˆš3<1<1+√31 - √3 < 1 < 1 + √3, our critical point x=1x=1 is within the domain. Hence, it is a candidate for the maximum. Now it's time to test if it's the maximum by either the first or second derivative test. It is up to you which one you decide to use. Let's look into the first derivative test.

The First Derivative Test

The first derivative test helps us determine whether a critical point is a local maximum, a local minimum, or neither. We analyze the sign of the derivative around the critical point. If the derivative changes from positive to negative at the critical point, we have a local maximum. If it changes from negative to positive, we have a local minimum. If the sign doesn't change, it's neither.

We know that the critical point is at x=1x=1. Let's examine the sign of fβ€²(x)=(2βˆ’2x)/((2+2xβˆ’x2)βˆ—ln(2))f'(x) = (2 - 2x) / ((2+2x-x^2) * ln(2)) around x=1x=1.

  1. For x<1x < 1: Let's pick a value in the domain less than 1, say x=0x = 0. Then fβ€²(0)=(2βˆ’2βˆ—0)/((2+2βˆ—0βˆ’02)βˆ—ln(2))=2/(2βˆ—ln(2))>0f'(0) = (2 - 2*0) / ((2+2*0-0^2) * ln(2)) = 2 / (2 * ln(2)) > 0. The derivative is positive. This means the function is increasing before x=1.
  2. For x>1x > 1: Let's pick a value in the domain greater than 1, say x=1.5x = 1.5. Then fβ€²(1.5)=(2βˆ’2βˆ—1.5)/((2+2βˆ—1.5βˆ’1.52)βˆ—ln(2))=βˆ’1/(2.75βˆ—ln(2))<0f'(1.5) = (2 - 2*1.5) / ((2+2*1.5-1.5^2) * ln(2)) = -1 / (2.75 * ln(2)) < 0. The derivative is negative. This means the function is decreasing after x=1.

Since the derivative changes from positive to negative at x=1x=1, we have a local maximum at x=1x=1. Since this is the only critical point in the domain, it must be the global maximum.

Finding the Maximum Value

Now that we've found the x-value that gives us the maximum, we need to find the y-value, the actual maximum value of the function. We do this by plugging the x-value (which is 1) back into the original function. Our function is f(x)=log2(2+2xβˆ’x2)βˆ’2f(x) = log_2 (2+2x-x^2) - 2.

So, we have: f(1)=log2(2+2(1)βˆ’(1)2)βˆ’2=log2(2+2βˆ’1)βˆ’2=log2(3)βˆ’2f(1) = log_2 (2 + 2(1) - (1)^2) - 2 = log_2 (2 + 2 - 1) - 2 = log_2(3) - 2. Because log2(3)βˆ’2log_2(3) - 2 can be simplified as log2(3)βˆ’log2(4)=log2(3/4)log_2(3) - log_2(4) = log_2(3/4). So the maximum value of the function is log2(3)βˆ’2log_2(3) - 2, which is approximately -0.415.

Conclusion

So, guys, we did it! We found the maximum point of the function f(x)=log2(2+2xβˆ’x2)βˆ’2f(x) = log_2 (2+2x-x^2) - 2. The x-value where the function reaches its maximum is x=1x = 1, and the maximum value of the function is log2(3)βˆ’2log_2(3) - 2. Remember the steps: determine the domain, find the critical points, use the first derivative test (or the second derivative test) to confirm the maximum, and calculate the maximum value by substituting the critical point value into the original function. Well done everyone!