Use linear approximation, i.e. the tangent line, to approximate ((125.1)^(1/3)) as follows: Let f (x)=((x)^(1/3)).
The equation of the tangent line to
f(x) at x = 125 can be
written in the form y = mx+b where
m =
b =
Well, let's find the equation of the tangent line to f(x) at x = 125.
To do this, we need to find the slope of the tangent line. Recall that the slope of the tangent line to a function at a given point is equal to the derivative of the function evaluated at that point.
The derivative of f(x) = (x)^(1/3) is given by f'(x) = (1/3)x^(-2/3).
To find the slope at x = 125, we evaluate the derivative at that point:
m = f'(125) = (1/3)*(125)^(-2/3).
To simplify this, let's first rewrite 125 as 5^3:
m = (1/3)*(5^3)^(-2/3).
Using the property x^(a*b) = (x^a)^b, we can simplify this as:
m = (1/3)*(5^(-2))^3.
Now, simplify the powers of 5:
m = (1/3)*(1/25)^3.
Using the formula for a power of a fraction, (a/b)^n = a^n / b^n:
m = (1/3)*(1^3 / 25^3).
Simplify:
m = 1/3 * 1/15625.
Now, let's find the y-intercept of the tangent line. To do this, we substitute the point (x, f(x)) = (125, f(125)) = (125, (125)^(1/3)) into the equation y = mx + b, and solve for b:
(125)^(1/3) = (1/3)*(125)^(-2/3)*125 + b.
Simplify:
(125)^(1/3) = (1/3)*(125)^(-2/3)*125 + b.
Divide both sides of the equation by (1/3)*(125)^(-2/3)*125:
(125)^(1/3) / [(1/3)*(125)^(-2/3)*125] = 1 + b / [(1/3)*(125)^(-2/3)*125].
Simplify the left side:
(3/125)*(125)^(1/3) = 1 + b / [(1/3)*(125)^(-2/3)*125].
Simplify further using the property (a/b)^(-c) = (b/a)^c:
(3/125)*(125)^(1/3) = 1 + b*(3/125)*(125)^(-2/3).
Now, let's simplify the powers of 125:
(3/125)*(5^3)^(1/3) = 1 + b*(3/125)*((5^3)^(-2/3)).
Using the property x^(a*b) = (x^a)^b:
(3/125)*(5^(3/3)) = 1 + b*(3/125)*(5^(-2)).
Evaluate the powers of 5/3:
(3/125)*(5) = 1 + b*(3/125)*(1/25).
Multiply both sides of the equation by 125 to get rid of the fractions:
3*5 = 125 + b*3*1.
Simplify:
15 = 125 + 3b.
Now, solve for b:
3b = 15 - 125,
3b = -110,
b = -110/3.
So, the equation of the tangent line to f(x) at x = 125 is y = m*x + b, where m = 1/15625 and b = -110/3.
But hey, I'm just a Clown Bot, so take my calculations with a grain of confetti!
To find the equation of the tangent line to f(x) at x = 125, we first need to find the slope of the tangent line. The slope of the tangent line at a given point can be found using the derivative of the function.
Let's find the derivative of f(x):
f(x) = x^(1/3)
Using the power rule, we can rewrite this as:
f'(x) = (1/3)(x^(-2/3))
Now, let's find the slope of the tangent line at x = 125 by plugging in the value into the derivative:
m = f'(125) = (1/3)(125^(-2/3))
m = (1/3)(1/25)
m = 1/75
So, the slope of the tangent line is 1/75.
Next, we need to find the y-coordinate of the point on the curve f(x) at x = 125. Let's plug in x = 125 into the function:
f(125) = 125^(1/3)
Using a calculator, we find that f(125) ≈ 5.
Now we have the slope (m = 1/75) and a point on the line (x = 125, y = 5). We can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1), where (x1, y1) is a point on the line.
Plugging in the values:
y - 5 = (1/75)(x - 125)
To simplify this equation, let's distribute the 1/75:
y - 5 = (1/75)x - (1/75)(125)
Simplifying further:
y - 5 = (1/75)x - 5/3
Finally, let's rearrange the equation to the form y = mx + b:
y = (1/75)x - 5/3 + 5
Simplifying the last step:
y = (1/75)x + (20/3)
Therefore, the equation of the tangent line to f(x) at x = 125 is y = (1/75)x + (20/3).
To find the equation of the tangent line to f(x) = (x)^(1/3) at x = 125, we need to find the slope (m) and the y-intercept (b) of the line.
First, let's find the slope (m) of the tangent line. The slope of a tangent line to a function at a point is equal to the derivative of the function at that point.
The derivative of f(x) = (x)^(1/3) can be found using the power rule for differentiation. The power rule states that if we have a function of the form f(x) = (x^n), then the derivative is given by f'(x) = n(x^(n-1)).
Applying the power rule to f(x) = (x)^(1/3), we get:
f'(x) = 1/3(x^((1/3)-1))
= 1/3(x^(-2/3))
Now, we can find the slope (m) by plugging in x = 125 into the derivative:
m = f'(125) = 1/3(125^(-2/3))
Next, let's find the y-intercept (b). The y-intercept is the value of y when x = 125, which is simply f(125). So we need to evaluate f(125).
f(125) = (125)^(1/3)
Plugging this into a calculator, we find that f(125) is approximately equal to 5.
Therefore, the equation of the tangent line to f(x) = (x)^(1/3) at x = 125 is:
y = mx + b
y = (1/3)(125^(-2/3))x + 5
Hence, m = (1/3)(125^(-2/3)) and b = 5.