se differential, i.e., linear approximation, to approximate (8.4)^(1/3) as follows:
Let f(x)=(x )^(1/3). The linear approximation to f(x) at x=8 can be written in the form y=mx+b where m is: and where b is:
Using this, we find our approximation for (8.4)^(1/3) is
y = x^(1/3)
dy/dx = (1/3) x^-(1/3)
at x = 8
y(8) = 2
dy/dx = (1/3)/2
y(x+h) = y(x) + h dy/dx
y(8.4) = 2 + .4 (1/3)/2
To find the linear approximation of \( f(x) = x^{1/3} \) at \( x = 8 \), we first need to find the derivative of \( f(x) \) with respect to \( x \).
The derivative of \( f(x) \) can be found using the power rule:
\[ f'(x) = \frac{1}{3} \cdot x^{-2/3} \]
Now let's find the slope of the linear approximation. The slope, \( m \), is equal to the value of the derivative at \( x = 8 \):
\[ m = f'(8) = \frac{1}{3} \cdot 8^{-2/3} \]
To find the y-intercept, \( b \), we substitute the coordinates of the point \( (8, f(8)) \) into the equation \( y = mx + b \). Plug in \( x = 8 \) and \( y = f(8) \):
\[ f(8) = 8^{1/3} \]
Simplify:
\[ f(8) = 2 \]
Now we can solve for \( b \):
\[ 2 = m \cdot 8 + b \]
Substitute the value of \( m \) we found earlier:
\[ 2 = \frac{1}{3} \cdot 8^{-2/3} \cdot 8 + b \]
Simplify:
\[ 2 = \frac{1}{3} \cdot 2 + b \]
\[ 2 = \frac{2}{3} + b \]
So, the y-intercept is:
\[ b = 2 - \frac{2}{3} = \frac{4}{3} \]
Therefore, the linear approximation of \( f(x) = x^{1/3} \) at \( x = 8 \) can be written as:
\[ y = mx + b = \left(\frac{1}{3} \cdot 8^{-2/3}\right) \cdot x + \frac{4}{3} \]
To find the approximation for \( (8.4)^{1/3} \), we can plug in \( x = 8.4 \) into the linear approximation equation:
\[ y = \left(\frac{1}{3} \cdot 8^{-2/3}\right) \cdot 8.4 + \frac{4}{3} \]
Simplify and evaluate:
\[ y = \left(\frac{1}{3} \cdot \left(\frac{1}{2}\right)^{2/3}\right) \cdot 8.4 + \frac{4}{3} \]
\[ y \approx 2.670047 \]
Therefore, the approximation for \( (8.4)^{1/3} \) using the linear approximation is approximately 2.670047.