Find the equation of the line tangent to the curve defined by f(x)=x2+1x√ at the point (1,2) . Show all your work. If limit notation is needed in your response, use the following example as a guide: limx→1(x2+2) should be written as lim x→1 (x2+2) .
To find the equation of the line tangent to the curve, we need to find the derivative of the function f(x) = x^2 + 1/x^(1/2).
Using the power rule, the derivative of x^2 is 2x, and the derivative of 1/x^(1/2) (which can also be written as x^(-1/2)) is -(1/2)x^(-3/2). Therefore, the derivative of f(x) is:
f'(x) = 2x - (1/2)x^(-3/2)
To find the slope of the tangent line at the point (1,2), substitute x=1 in the derivative:
f'(1) = 2(1) - (1/2)(1)^(-3/2)
= 2 - (1/2)(1)
= 2 - 1/2
= 3/2.
So, the slope of the tangent line at (1,2) is 3/2.
Now, we have the slope of the tangent line and a point on the line (1,2). We can use the point-slope form of the equation of a line:
y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Plugging in the values, we get:
y - 2 = (3/2)(x - 1).
Simplifying, we have:
y - 2 = (3/2)x - 3/2.
Bringing the constant term to the other side, we get:
y = (3/2)x - 3/2 + 2
= (3/2)x + 1/2.
Therefore, the equation of the line tangent to the curve f(x) = x^2 + 1/x^(1/2) at the point (1,2) is y = (3/2)x + 1/2.
To find the equation of the tangent line to the curve defined by f(x) = x^2 + 1/x√ at the point (1,2), we first need to find the derivative of the function f(x) with respect to x.
Step 1: Find the derivative f'(x):
To find the derivative of f(x), we can use the power rule and the chain rule.
First, let's differentiate the term x^2:
d/dx (x^2) = 2x
Next, we differentiate 1/x√:
Let u = 1/x√
Applying the chain rule:
d/dx (1/x√) = d/du (1/u) * du/dx
For d/du (1/u), we can use the power rule:
d/du (1/u) = -1/u^2 = -1/(1/x√)^2 = -x^2
For du/dx, we differentiate u = 1/x√ with respect to x:
du/dx = -1/2 * x^(-3/2) = -1/(2x^3/2)
Putting it all together:
d/dx (1/x√) = -x^2 * -1/(2x^3/2) = x^2/(2x^3/2) = x^2 * 2x^(3/2) = 2x^(5/2)
Now, let's find the derivative of f(x) using the sum rule:
f'(x) = d/dx (x^2) + d/dx (1/x√)
= 2x + 2x^(5/2)
Step 2: Find the slope of the tangent line:
To find the slope of the tangent line at the point (1,2), we substitute x = 1 into f'(x):
slope = f'(1) = 2(1) + 2(1)^(5/2) = 2 + 2 = 4
Step 3: Find the equation of the tangent line:
Using the point-slope form of a line, we have:
y - y₁ = m(x - x₁)
Plugging in the values x₁ = 1, y₁ = 2, and slope m = 4:
y - 2 = 4(x - 1)
Expanding and simplifying:
y - 2 = 4x - 4
y = 4x - 2
Therefore, the equation of the tangent line to the curve f(x) = x^2 + 1/x√ at the point (1,2) is y = 4x - 2.
To find the equation of the tangent line at a given point on a curve, we can follow these steps:
Step 1: Find the derivative of the function f(x).
Step 2: Evaluate the derivative at the given point to find the slope of the tangent line.
Step 3: Use the slope and the given point to write the equation of the tangent line in point-slope form.
Step 4: Simplify the equation, if necessary, to a more standard form.
Let's apply these steps to the given problem:
Step 1: Find the derivative of f(x):
To find the derivative of f(x) = x^2 + 1/x√, we will use the power rule and the product rule.
Differentiating x^2, we get 2x.
Differentiating 1/(x√), we get -1/2x^2*√x.
So, the derivative of f(x), denoted by f'(x), is:
f'(x) = 2x - 1/(2x^2*√x)
Step 2: Evaluate the derivative at the given point (1,2):
To find the slope of the tangent line at the point (1,2), substitute x = 1 into the derivative f'(x):
f'(1) = 2(1) - 1/(2(1)^2*√1)
= 2 - 1/2
= 3/2
Therefore, the slope of the tangent line is 3/2.
Step 3: Use the slope and the given point to write the equation of the tangent line in point-slope form:
Using the point-slope form:
y - y1 = m(x - x1)
Where (x1, y1) is the given point, and m is the slope.
The given point is (1,2) and the slope is 3/2. So we have:
y - 2 = (3/2)(x - 1)
Step 4: Simplify the equation to a more standard form:
Multiply through by 2 to eliminate the fraction:
2y - 4 = 3(x - 1)
Simplifying further, we get:
2y - 4 = 3x - 3
Finally, rearranging the equation to the standard form, we have:
3x - 2y = 1
Therefore, the equation of the tangent line to the curve defined by f(x) = x^2 + 1/x√ at the point (1,2) is 3x - 2y = 1.