Use linear approximation, i.e. the tangent line, to approximate (the 3 is inside the v part if you know what I mean, it's not 3 times sqrt) 3√125.2 as follows: Let f(x) = 3√x. The equation of the tangent line to f(x) at x=125 can be written in the form y=mx+b where m is ____ and where b is _____
Using this, we find our approximation for 3√125.2 is _____
(again, all the 3√, the 3 is inside the v, not 3 times sqrt)
So, m and b are?
Nevermind, got it.
To find the equation of the tangent line to the function f(x) = 3√x at x=125, we need to calculate the slope and the y-intercept of the line.
1. Calculate the derivative of f(x) with respect to x:
f'(x) = (d/dx) (3√x) = (1/2)*(3√x)^(-1/2) = 3/(2√x)
2. Substitute x=125 into f'(x) to find the slope:
m = f'(125) = 3/(2√125) = 3/(2*5) = 3/10
3. Find the y-coordinate of the point on the function f(x) using x=125:
f(125) = 3√125 = 3√(5^3) = 3*5 = 15
Since the tangent line passes through the point (125, 15) and has slope m = 3/10, the equation of the tangent line, in the form y = mx + b, can be written as y = (3/10)x + b.
To find b, substitute the coordinates (125, 15) into the equation:
15 = (3/10)*125 + b
Simplifying the equation:
15 = 37.5 + b
b = 15 - 37.5
b = -22.5
Therefore, the equation of the tangent line is y = (3/10)x - 22.5.
Now, to approximate 3√125.2 using this tangent line approximation:
1. Substitute x = 125.2 into the equation of the tangent line:
y ≈ (3/10)*(125.2) - 22.5
y ≈ 37.56 - 22.5
y ≈ 15.06
Hence, the approximation for 3√125.2 using linear approximation is approximately 15.06.
In general,
f(x) ≅ f(x0) + f'(x0)*(x-x0)
For
f(x) = ∛x
f(x) ≅ ∛x0 + (x-x0)/(3∛(x0²))
Let x0=125, x=125.2
f(125.2) = f(125) + (125.2-125)/(3∛(125²))
=5 + (0.2)/(3*25)
=5.0267
Check: 5.0267³ = 125.2001...