Use a linear approximation (or differentials) to estimate the given number
(2.001)^5
Let y=x5,
y'=5x4.
y(x+h)=y(x)+y'(2)*h (approximately)
Therefore, an approximation to y(2.001) is
y(2.001)
=y(2)+y'(2)*0.001
=2^5 + 5(2)^4*0.001 (approximately)
approximate 4 square root of 15.8
To estimate (2.001)^5 using linear approximation or differentials, we can start by finding the equation of the tangent line at a nearby point, which is easier to work with.
Let's choose a nearby point to approximate from, such as x = 2.
To find the equation of the tangent line at x = 2, we'll need the slope and the y-intercept.
1. Find the slope (m):
The slope of the tangent line can be found by taking the derivative of the function.
d/dx (x^5) = 5x^4
Evaluate the derivative at x = 2 to find the slope.
m = 5(2)^4 = 80
2. Find the y-intercept (b):
To find the y-intercept, substitute the point (2, (2)^5) into the equation.
(2)^5 = 32
b = 32
Now, we have the equation of the tangent line:
y = mx + b
y = 80x + 32
Using this equation, we can estimate the value of (2.001)^5 by substituting x = 0.001 into the equation:
y = 80(0.001) + 32
y = 0.08 + 32
y ≈ 32.08
Therefore, the estimate of (2.001)^5 using linear approximation is approximately 32.08.
To estimate (2.001)^5 using a linear approximation, we need to find the equation of a tangent line to the function f(x) = x^5 at the point x = 2.
First, we calculate the derivative of f(x) which is given by f'(x) = 5x^4.
Next, we evaluate f'(2) to find the slope of the tangent line at x = 2. Plugging in x = 2 into f'(x) gives us f'(2) = 5(2)^4 = 80.
Now, we have the slope of the tangent line and a known point on the line (2, f(2)) = (2, 2^5 = 32).
Using the point-slope form of a linear equation, y - y1 = m(x - x1), we can write the equation for the tangent line as y - 32 = 80(x - 2).
To estimate (2.001)^5, we substitute x = 2.001 into the equation of the tangent line:
y - 32 = 80(2.001 - 2)
y - 32 = 80 * 0.001
y - 32 = 0.08
y = 32.08
Therefore, using the linear approximation, we estimate that (2.001)^5 is approximately equal to 32.08.