Use linear approximation, i.e. the tangent line, to approximate cube root of 27.05 as follows. Let f(x) = cube root of x and find the equation of the tangent line to f(x) at x = 27 in the form y = mx+b.
Note: The values of m and b are rational numbers which can be computed by hand. You need to enter expressions which give m and b exactly. You may not have a decimal point in the answers to either of these parts.
A. m = ?
B. b= ?
C.Using these values, find the approximation. Cube root of 27.05 = ?
this is just algebra I, after finding the slope of the tangent line...
since y = x^(1/3)
y' = 1/3 x^-(2/3)
y(27) = 3
y'(27) = 1/9
So, now you have a point and a slope, so the line is
y-3 = 1/9 (x-27)
I'll let you massage the equation.
Now you can use that line to find y(27.05)
Or, you can note that
dy = 1/9 dx
and dx = 0.05,
so you can add dy to y to approximate y(27.05)
Hint, using the line or using the differentials will give the same answer.
To find the equation of the tangent line to f(x) at x = 27, we first need to find the derivative of f(x).
Let's start by finding f'(x) using the power rule for differentiation:
f(x) = x^(1/3)
f'(x) = (1/3)*x^(-2/3)
Evaluate f'(x) at x = 27 to find the slope of the tangent line:
f'(27) = (1/3)*(27^(-2/3))
To find m (the slope of the tangent line), we need to evaluate this expression exactly:
m = (1/3)*(27^(-2/3))
Simplifying this expression gives:
m = (1/3)*(1/3)
m = 1/9
So, the value of m is 1/9.
To find b (the y-intercept of the tangent line), substitute the point (27, f(27)) = (27, 27^(1/3)) into the equation y = mx + b:
27^(1/3) = (1/9)*27 + b
To find b, solve this equation for b:
27^(1/3) = 3/3 + b
27^(1/3) = 1 + b
b = 27^(1/3) - 1
So, the value of b is 27^(1/3) - 1.
Now, to find the approximation for the cube root of 27.05, substitute x = 27.05 into the equation of the tangent line:
y = (1/9)*x + (27^(1/3) - 1)
Evaluate this expression to find the approximate value:
cube root of 27.05 = (1/9)*27.05 + (27^(1/3) - 1)
Simplifying this expression will give you the numerical approximation.
To approximate the cube root of 27.05 using linear approximation, we first need to find the equation of the tangent line to the function f(x) = cube root of x at x = 27.
Step 1: Find the slope of the tangent line
To find the slope, we can take the derivative of the function f(x) = cube root of x. The derivative of f(x) can be found using the power rule of differentiation.
f'(x) = (1/3)x^(-2/3)
Evaluate f'(x) at x = 27:
f'(27) = (1/3)(27)^(-2/3)
Simplifying,
f'(27) = (1/3)(1/3^2) = (1/3)(1/9) = 1/27
Therefore, the slope of the tangent line, m, is 1/27.
Step 2: Find the y-intercept of the tangent line
To find the y-intercept, we need the point (x,y) that lies on the tangent line. In this case, the point is (27,f(27)).
f(27) = cube root of 27 = 3
So, the point (27, 3) lies on the tangent line.
Using the slope-intercept form of a line (y=mx+b), we can substitute the slope and the point (27, 3) to find the y-intercept, b.
3 = (1/27)(27) + b
Simplifying,
3 = 1 + b
b = 2
Therefore, the y-intercept of the tangent line, b, is 2.
Now we have the equation of the tangent line:
y = (1/27)x + 2
To find the approximation of the cube root of 27.05, we need to substitute x = 27.05 into the equation of the tangent line and solve for y.
y = (1/27)(27.05) + 2
Calculating this expression,
y ≈ 1.00204 + 2
y ≈ 3.00204
Therefore, the approximation of the cube root of 27.05 is approximately 3.00204.
A. m = 1/27
B. b = 2
C. The approximation of the cube root of 27.05 is approximately 3.00204.