Tangent Line Equation & Approximation: F(x) = Cbrt(-128x^3+640)

Hey guys! Let's dive into a fun math problem involving finding the tangent line to a function and using it for approximation. Today, we're tackling the function f(x) = \sqrt[3]{-128x^3 + 640} and we're going to find the tangent line at x = 1. We'll then use this tangent line to approximate the value of f(1.1) and see how accurate our approximation is. So, buckle up, and let’s get started!

1. Finding the Tangent Line Equation

To find the equation of the tangent line, we need two key ingredients: the slope of the tangent line at the given point and a point on the line. Remember the tangent line equation formula: y = mx + b, where m is the slope and b is the y-intercept. To calculate the slope, we need to compute the derivative of the function f(x) and evaluate it at x = 1. This will give us the instantaneous rate of change of the function at that specific point, which is exactly what the slope of the tangent line represents. Also, we should find the point of tangency, which is simply the value of f(1). This value, along with x=1, will be our (x1, y1) point for constructing the equation using the point-slope form later on.

1.1. Calculate the Derivative, f'(x)

First, let's find the derivative of f(x) = \sqrt[3]{-128x^3 + 640}. We can rewrite this as f(x) = (-128x^3 + 640)^(1/3). This will make it easier to apply the power rule and the chain rule. Remember the chain rule: if you have a composite function, the derivative is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Also, don't forget that the power rule states that the derivative of x^n is nx^(n-1).

Applying the chain rule and the power rule, we get:

f'(x) = (1/3)(-128x^3 + 640)^(-2/3) * (-384x^2)

Let's simplify this a bit:

f'(x) = -128x^2 / (-128x^3 + 640)^(2/3)

This is our derivative! It might look a little intimidating, but don't worry, we'll plug in x = 1 soon, which will simplify things. Derivatives are super important in calculus because they tell us how a function is changing at any given point. The derivative we just found tells us the slope of the tangent line to the graph of f(x) at any value of x.

1.2. Evaluate f'(1) to Find the Slope (m)

Now, we need to find the slope of the tangent line at x = 1. We do this by plugging x = 1 into our derivative, f'(x):

f'(1) = -128(1)^2 / (-128(1)^3 + 640)^(2/3)

f'(1) = -128 / (-128 + 640)^(2/3)

f'(1) = -128 / (512)^(2/3)

Since 512 is 8 cubed (8^3), (512)^(2/3) is (8^2) which equals 64:

f'(1) = -128 / 64

f'(1) = -2

So, the slope of the tangent line at x = 1 is -2. This means that at the point where x = 1, the function f(x) is decreasing at a rate of 2 units for every 1 unit increase in x. This is a crucial piece of information for constructing the tangent line equation. We now know our m value in the tangent line equation y = mx + b.

1.3. Find the Point of Tangency (x₁, y₁)

To write the equation of the tangent line, we also need a point on the line. We know the tangent line touches the graph of f(x) at x = 1. So, the x-coordinate of our point is 1. To find the y-coordinate, we need to evaluate the original function, f(x), at x = 1:

f(1) = \sqrt[3]{-128(1)^3 + 640}

f(1) = \sqrt[3]{-128 + 640}

f(1) = \sqrt[3]{512}

f(1) = 8

Thus, the point of tangency is (1, 8). This point lies both on the original function f(x) and on the tangent line we are trying to find. This is the essential connection point that allows us to define the tangent line.

1.4. Write the Tangent Line Equation

Now we have everything we need! We have the slope, m = -2, and a point on the line, (1, 8). We can use the point-slope form of a line equation: y - y₁ = m(x - x₁). Plugging in our values, we get:

y - 8 = -2(x - 1)

Let's simplify this to slope-intercept form (y = mx + b) to make it look cleaner:

y - 8 = -2x + 2

y = -2x + 10

So, the equation of the tangent line to f(x) at x = 1 is y = -2x + 10. This line is the best linear approximation of the function f(x) near the point x = 1. It's a straight line that skims the curve of f(x) at that point, giving us a simple way to estimate the function's value for nearby x values.

2. Approximating f(1.1) Using the Tangent Line

Okay, now for the fun part: using our tangent line to approximate the value of f(1.1). The idea here is that near the point of tangency (x = 1), the tangent line is a pretty good approximation of the original function. So, instead of calculating the cube root directly, we can just plug x = 1.1 into the tangent line equation.

To approximate f(1.1), we substitute x = 1.1 into the tangent line equation, y = -2x + 10:

y = -2(1.1) + 10

y = -2.2 + 10

y = 7.8

Therefore, using the tangent line approximation, we estimate that f(1.1) is approximately 7.8. This is a quick and easy way to get an estimate of the function's value without having to deal with the cube root calculation. Tangent line approximations are super useful when evaluating a function directly is difficult or computationally expensive.

3. Calculating the Error

Now, let's see how accurate our approximation is. To do this, we need to calculate the actual value of f(1.1) and then compare it to our approximation. The error is the absolute difference between the actual value and the approximation.

3.1. Find the Actual Value of f(1.1)

To find the actual value, we plug x = 1.1 into the original function, f(x) = \sqrt[3]{-128x^3 + 640}:

f(1.1) = \sqrt[3]{-128(1.1)^3 + 640}

f(1.1) = \sqrt[3]{-128(1.331) + 640}

f(1.1) = \sqrt[3]{-170.368 + 640}

f(1.1) = \sqrt[3]{469.632}

Using a calculator, we find that:

f(1.1) ≈ 7.766 (rounded to three decimal places)

So, the actual value of f(1.1) is approximately 7.766. This is the true value of the function at x = 1.1, and we'll use it to determine the accuracy of our linear approximation.

3.2. Calculate the Absolute Error

The error is the absolute difference between the approximated value and the actual value. We approximated f(1.1) as 7.8, and the actual value is approximately 7.766. Therefore, the error is:

Error = |Approximation - Actual Value|

Error = |7.8 - 7.766|

Error = |0.034|

Error = 0.034

The error in our approximation is approximately 0.034. This is a relatively small error, which means our tangent line approximation was pretty good! This illustrates the power of linear approximation: it allows us to estimate function values with reasonable accuracy using a simple linear equation.

4. Conclusion

Alright, guys, we made it! We successfully found the equation of the tangent line to the graph of f(x) = \sqrt[3]{-128x^3 + 640} at x = 1, which was y = -2x + 10. We then used this tangent line to approximate f(1.1), getting an approximation of 7.8. Finally, we calculated the actual value of f(1.1) (approximately 7.766) and found that the error in our approximation was only 0.034.

This exercise demonstrates the usefulness of tangent lines in approximating function values. When dealing with complex functions, tangent lines provide a simple and effective way to estimate values near a specific point. Keep practicing these techniques, and you'll become a pro at using calculus for approximations! Remember that the closer you are to the point of tangency, the better the approximation will be. This is because the tangent line is the best linear representation of the function at that specific point.