Use linear approximation, i.e. the tangent line, to approximate \sqrt[3] { 7.9 } as follows:

The equation of the tangent line to f(x) at x = 8 can be written in the form y =
Using this, we find our approximation for \sqrt[3] {7.9} is

It appears you are trying to approximate ∛7.9

So, if y = ∛x, then
dy = 1/(3∛(x^2)) dx
at x=8, y=2, so if dx = -0.1,

dy = 1/(3∛64) (-0.1)
= -1/240 = -.0042

So, ∛7.9 ≈ 1.9958

check: ∛7.9 is 1.9916
The estimate is expected to be a bit high, since ∛x is concave down.

Oh, yeah. The tangent line is

y-2 = 1/24 (x-8)

To find the equation of the tangent line to the function f(x) at x = 8, we first need to find the derivative of f(x).

Let's start by finding the derivative of f(x) = √[3]{x}.

Using the power rule for differentiation, the derivative of f(x) with respect to x is:

f'(x) = (1/3) * (x^(-2/3))

Next, we evaluate f'(x) at x = 8 to find the slope of the tangent line.

f'(8) = (1/3) * (8^(-2/3))
= (1/3) * (1/4)
= 1/12

So, the slope of the tangent line is 1/12.

Now we have the slope and a point (8, f(8)) = (8, √[3]{8}) on the line.

Using the point-slope form of the equation of a line,

y - y1 = m(x - x1),

we can plug in the values to find the equation of the tangent line:

y - √[3]{8} = (1/12)(x - 8)

Simplifying:

y - √[3]{8} = (1/12)x - 2/3

Rearranging to the desired form:

y = (1/12)x - 2/3 + √[3]{8}

Therefore, using this equation, our approximation for √[3]{7.9} is

y = (1/12)(7.9) - 2/3 + √[3]{8}

To approximate \sqrt[3]{7.9} using linear approximation, we start by considering the function f(x) = \sqrt[3]{x}. We want to find the equation of the tangent line to f(x) at x = 8.

To find the equation of a tangent line, we need to find the slope and the y-intercept of the line.

The slope of the tangent line can be found by taking the derivative of f(x) with respect to x. The derivative of f(x) = \sqrt[3]{x} is given by:

f'(x) = 1/(3 * \sqrt[3]{x^2})

To find the slope at x = 8, we substitute x = 8 into the derivative:

f'(8) = 1/(3 * \sqrt[3]{8^2})
= 1/(3 * \sqrt[3]{64})
= 1/(3 * 4)
= 1/12

So the slope of the tangent line at x = 8 is 1/12.

Next, we need to find the y-intercept of the tangent line. To do this, we substitute the coordinates (8, f(8)) = (8, \sqrt[3]{8}) into the equation y = mx + b, where m is the slope and b is the y-intercept.

Using (x, y) = (8, \sqrt[3]{8}), we have:

\sqrt[3]{8} = (1/12) * 8 + b
\sqrt[3]{8} = 2/3 + b

To isolate b, we subtract 2/3 from both sides:

b = \sqrt[3]{8} - 2/3

So the y-intercept of the tangent line is \sqrt[3]{8} - 2/3.

Putting it all together, the equation of the tangent line to f(x) at x = 8 is:

y = (1/12)x + (\sqrt[3]{8} - 2/3)

Using this equation, we can approximate \sqrt[3]{7.9} by substituting x = 7.9:

y = (1/12) * 7.9 + (\sqrt[3]{8} - 2/3)
= 7.9/12 + (\sqrt[3]{8} - 2/3)

Evaluating this expression will give us the approximation for \sqrt[3]{7.9}.