Use differential, i.e., linear approximation, to approximate (125.4^(1/3)) as follows:
Let f(x)=x^(1/3) . The linear approximation to f(x) at x=125 can be written in the form y=mx+b where m is:
m=1/75
b=?
How do I find b?
f(x) = x^(1/3) = 125^(1/3) = 5
f'(x) = (1/3) x^(-2/3) = 1/3 / x^(2/3)
for x = 125, this is (1/3)/ 25 = 1/75 by the way
f(x+delta x) = f(x) + delta x * f'(x)
f(125+.4) = 5 + .4 *(1/75)
so b = 5 and m = (1/75)
To find the value of b in the linear approximation y = mx + b, we need to determine the y-value when x = 125.
First, we have the function f(x) = x^(1/3). To find the linear approximation, we use the concept of the tangent line at x = 125.
The derivative of f(x) can be found using the power rule.
f'(x) = (1/3) * x^(-2/3)
Substituting x = 125 into f'(x), we get:
f'(125) = (1/3) * 125^(-2/3)
Simplifying this expression gives us:
f'(125) = (1/3) * (1/125^(2/3))
Next, recall that the slope, m, of the linear approximation is equal to f'(125).
So, m = f'(125) = (1/3) * (1/125^(2/3))
Now we have determined the value of m. To find b, we need a point on the line. We can use the original function f(x) at x = 125 as the point.
f(125) = 125^(1/3)
This gives us the y-value or f(125) when x = 125.
Finally, we substitute the values of m and (125^(1/3)) into the equation y = mx + b and solve for b:
(125^(1/3)) = m * 125 + b
Substituting the value of m and (125^(1/3)), we have:
(125^(1/3)) = (1/3) * (1/125^(2/3)) * 125 + b
Simplifying this equation gives us the value of b.
Therefore, to find b, evaluate the expression:
b = (125^(1/3)) - (1/3) * (1/125^(2/3)) * 125