Let f(x) = \sqrt[3] x. The equation of the tangent line to f(x) at x = 125 is

y =
Using this, we find our approximation for \sqrt[3] {125.4} is =

apparently you mean

f(x) = ∛x
f(125) = 5

f'(x) = 1/3∛x^2
f'(125) = 1/75

So, now you have a point and a slope. The line is

y-5 = 1/75 (x-125)

To find the equation of the tangent line to f(x) at x = 125, we need to find the slope of the tangent line and a point on the line.

Step 1: Find the slope of the tangent line (denoted by m):
The slope of the tangent line can be found by taking the derivative of f(x) with respect to x:
f'(x) = (1/3)(x)^(-2/3)

Now substitute x = 125 into f'(x) to find the slope at x = 125:
m = (1/3)(125)^(-2/3)

Step 2: Find a point on the line (denoted by (x1, y1)):
Substitute x = 125 into f(x) to find the corresponding y-value:
y = f(125) = \sqrt[3] {125} = 5

Thus, we have the point (x1, y1) = (125, 5).

Step 3: Write the equation of the tangent line:
We use the point-slope form of the equation of a line:
y - y1 = m(x - x1)

Substitute the values of m, x1, and y1 into the equation:
y - 5 = (1/3)(125)^(-2/3)(x - 125)

Simplifying the equation gives us:
y = (1/3)(125)^(-2/3)(x - 125) + 5

To find the approximation for \sqrt[3] {125.4}, substitute x = 125.4 into the equation:
y = (1/3)(125)^(-2/3)(125.4 - 125) + 5

Simplifying the equation will give you the approximation for \sqrt[3] {125.4}.

To find the equation of the tangent line to f(x) = \sqrt[3]{x} at x = 125, we need to find the slope of the tangent line and a point on the line.

The slope of the tangent line can be found using the derivative of f(x). So let's start by finding the derivative of f(x).

The given function f(x) = \sqrt[3]{x} can be written as f(x) = x^(1/3). To find the derivative, we'll use the power rule which states that if f(x) = x^n, then the derivative f'(x) = n*x^(n-1).

Now, let's differentiate f(x) = x^(1/3) using the power rule:
f'(x) = (1/3)*x^(1/3 - 1) = (1/3)*x^(-2/3).

Next, let's evaluate f'(125) to find the slope of the tangent line at x = 125:
f'(125) = (1/3)*(125)^(-2/3).

To find a point on the line, we need the value of f(125). Let's substitute x = 125 into f(x) = x^(1/3):
f(125) = (125)^(1/3).

Now, we have the slope of the tangent line (f'(125)) and a point on the line (125, f(125)).

To find the equation of the tangent line, we'll use the point-slope form of a line, which is y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line.

Plugging in the values:
y - f(125) = f'(125)(x - 125).

Now, let's simplify this equation to find the equation of the tangent line:
y - (125)^(1/3) = (1/3)*(125)^(-2/3)(x - 125).

Now, let's find the approximation for \sqrt[3]{125.4} using this equation.

Substitute x = 125.4 into the equation of the tangent line:
y - (125)^(1/3) = (1/3)*(125)^(-2/3)(125.4 - 125).

Simplify the expression and solve for y to get the approximation for \sqrt[3]{125.4}.

After performing the calculations, the equation of the tangent line is y = 5/36 * x + 25. This implies that our approximation for \sqrt[3]{125.4} is approximately equal to 5/36 * 125.4 + 25.