Minima Problems: Tangent & Exponential Functions

October 15, 2025 by TextBrain Team 49 views

Hey guys! Let's dive into some cool problems involving finding the minimum values of functions. These types of questions pop up quite a bit in calculus, and mastering them can seriously boost your math skills. We're going to break down two specific problems: one involving a tangent function and another with an exponential. So, buckle up, and let’s get started!

Finding the Minimum Value of y = 66tan(x) - 132x + 33π + 7

So, our first mission is to find the smallest value this function hits on the stretch from -π/3 to π/3. This is where things get interesting, and we need to put on our calculus hats. Remember, finding the minimum or maximum value of a function usually involves looking at its critical points.

Understanding the Function

Before we jump into the calculations, let’s get a feel for what we're dealing with. Our function, y = 66tan(x) - 132x + 33π + 7, combines a tangent function with a linear term. The tangent function, tan(x), has a periodic, wave-like nature with vertical asymptotes, while the term -132x is a straight line sloping downwards. The constants 33π and 7 simply shift the entire graph up, but don't affect where the minimum occurs. So, we're essentially trying to balance the curves and lines to find the lowest point within our given interval.

Step 1: Finding the Derivative

The first move in finding the minimum (or maximum) of a function is to take its derivative. The derivative tells us about the function's slope at any given point. Critical points, where the function's slope is zero or undefined, are prime spots for potential minima or maxima. So let's find that derivative:

y = 66tan(x) - 132x + 33π + 7
y' = 66sec²(x) - 132

Remember that the derivative of tan(x) is sec²(x), and the derivative of cx (where c is a constant) is simply c. The constants 33π and 7 vanish because the derivative of a constant is zero.

Step 2: Setting the Derivative to Zero

Now comes the crucial part: we set the derivative equal to zero and solve for x. This will give us the x-values where the tangent line to the curve is horizontal, which are our potential minimum or maximum points:

66sec²(x) - 132 = 0

Let's simplify this equation:

66sec²(x) = 132
sec²(x) = 2

Now, recall that sec(x) is the reciprocal of cos(x), so sec²(x) = 1/cos²(x). Let’s rewrite the equation:

1/cos²(x) = 2
cos²(x) = 1/2

Taking the square root of both sides, we get:

cos(x) = ±√(1/2) = ±1/√2 = ±√2/2

Step 3: Solving for x

Now we need to find the values of x within our interval [-π/3, π/3] that satisfy cos(x) = ±√2/2. Remember your unit circle! Cosine corresponds to the x-coordinate on the unit circle.

cos(x) = √2/2 at x = π/4
cos(x) = -√2/2 at x = -π/4

Both π/4 and -π/4 fall nicely within our interval of [-π/3, π/3], so these are our critical points.

Step 4: Evaluating the Function at Critical Points and Endpoints

We've found our critical points, but we're not quite done yet. To find the absolute minimum on the interval, we need to check the value of the function at these critical points and at the endpoints of the interval, which are -π/3 and π/3. This is because the minimum could occur at a turning point (critical point) or at the very edge of our interval.

Let’s calculate:

y(-π/3) = 66tan(-π/3) - 132(-π/3) + 33π + 7 = 66(-√3) + 44π + 33π + 7 = -66√3 + 77π + 7
y(-π/4) = 66tan(-π/4) - 132(-π/4) + 33π + 7 = 66(-1) + 33π + 33π + 7 = -66 + 66π + 7
y(π/4) = 66tan(π/4) - 132(π/4) + 33π + 7 = 66(1) - 33π + 33π + 7 = 66 + 7 = 73
y(π/3) = 66tan(π/3) - 132(π/3) + 33π + 7 = 66√3 - 44π + 33π + 7 = 66√3 - 11π + 7

Step 5: Identifying the Minimum Value

Now comes the moment of truth! We need to compare the values we just calculated to see which one is the smallest. This might require a calculator since we have some irrational numbers involved.

y(-π/3) ≈ -66√3 + 77π + 7 ≈ -114.32 + 241.90 + 7 ≈ 134.58
y(-π/4) ≈ -66 + 66π + 7 ≈ -66 + 207.35 + 7 ≈ 148.35
y(π/4) = 73
y(π/3) ≈ 66√3 - 11π + 7 ≈ 114.32 - 34.56 + 7 ≈ 86.76

Comparing these values, it's clear that the smallest value is y(π/4) = 73.

Conclusion for the Tangent Function Problem

So, the minimum value of the function y = 66tan(x) - 132x + 33π + 7 on the interval [-π/3, π/3] is 73, and it occurs at x = π/4. Awesome! We’ve navigated through the derivatives, critical points, and endpoints to nail down the minimum. This process is a cornerstone of calculus, and you've just rocked it!

Finding the Minimum Point of y = (8x² - 40x + 40)e^(x+4)

Alright, let's switch gears and tackle another exciting problem! This time, we're looking for the minimum point of the function y = (8x² - 40x + 40)e^(x+4). Notice that we’re asked for the minimum point, which means we're looking for the x-coordinate where the function hits its minimum. This function involves a quadratic polynomial multiplied by an exponential, making it a classic example for applying the product rule and analyzing exponential behavior.

Understanding the Function

Before we dive into the calculations, let's take a moment to understand our function. We have a quadratic polynomial, (8x² - 40x + 40), which represents a parabola. This parabola is then multiplied by the exponential function e^(x+4). Exponential functions are fascinating because they grow (or decay) very rapidly. The product of these two functions creates a more complex behavior, but the exponential factor will play a significant role in determining the overall shape of the function, especially for large values of x.

Step 1: Finding the Derivative

Just like before, our first step towards finding the minimum point is to find the derivative of the function. Since we have a product of two functions, we'll need to use the product rule. The product rule states that if y = u(x)v(x), then y' = u'(x)v(x) + u(x)v'(x).

Let’s identify our u(x) and v(x):

u(x) = 8x² - 40x + 40
v(x) = e^(x+4)

Now, let's find their derivatives:

u'(x) = 16x - 40
v'(x) = e^(x+4) (Remember, the derivative of e^u is e^u * u', and in this case, the derivative of (x+4) is just 1)

Now we can apply the product rule:

y' = (16x - 40)e^(x+4) + (8x² - 40x + 40)e^(x+4)

Step 2: Simplifying the Derivative

Derivatives can often look a bit messy at first, so let's simplify ours. We can factor out the common factor of e^(x+4):

y' = e^(x+4) [(16x - 40) + (8x² - 40x + 40)]

Now, let's simplify the expression inside the brackets:

y' = e^(x+4) [8x² - 24x]

We can further simplify by factoring out 8x from the brackets:

y' = 8xe^(x+4) [x - 3]

Step 3: Setting the Derivative to Zero

Now, let's set the derivative equal to zero to find our critical points:

8xe^(x+4) [x - 3] = 0

This gives us three factors to consider: 8x, e^(x+4), and (x - 3). Let's analyze each one:

8x = 0 => x = 0
e^(x+4) = 0 => This factor will never be zero because the exponential function is always positive.
x - 3 = 0 => x = 3

So, our critical points are x = 0 and x = 3.

Step 4: Determining the Nature of the Critical Points

We've found our critical points, but we need to determine whether they correspond to a minimum, a maximum, or neither. There are a couple of ways to do this: we can use the second derivative test or analyze the sign of the first derivative around the critical points. Let’s use the first derivative test since it gives a more intuitive understanding of what's happening.

We'll create a sign chart for y' = 8xe^(x+4) [x - 3] by testing values in the intervals defined by our critical points:

Interval 1: x < 0 (Let's test x = -1)
- y'(-1) = 8(-1)e^(-1+4) [-1 - 3] = (-)(+)(-) = Positive
Interval 2: 0 < x < 3 (Let's test x = 1)
- y'(1) = 8(1)e^(1+4) [1 - 3] = (+)(+)(-) = Negative
Interval 3: x > 3 (Let's test x = 4)
- y'(4) = 8(4)e^(4+4) [4 - 3] = (+)(+)(+) = Positive

Step 5: Interpreting the Sign Chart

Here's what our sign chart tells us:

For x < 0, y' is positive, which means the function is increasing.
For 0 < x < 3, y' is negative, which means the function is decreasing.
For x > 3, y' is positive, which means the function is increasing.

This pattern indicates that:

At x = 0, the function changes from increasing to decreasing, so we have a local maximum.
At x = 3, the function changes from decreasing to increasing, so we have a local minimum.

Conclusion for the Exponential Function Problem

Therefore, the minimum point of the function y = (8x² - 40x + 40)e^(x+4) occurs at x = 3. We've successfully used the product rule, found critical points, and analyzed the sign of the derivative to pinpoint where the function bottoms out. Great job!

Final Thoughts

So there you have it, guys! We've tackled two fantastic problems involving finding minimum values. We’ve seen how to work with both trigonometric and exponential functions, and we've reinforced the importance of derivatives, critical points, and sign analysis. Remember, these techniques are powerful tools in calculus, and the more you practice, the more confident you'll become in using them. Keep up the awesome work, and happy problem-solving!