How do you find the equation of the line tangent to #f(x)=3/x^2# at x=2?

Question

Elijah Abram · Accepted Answer

#y=-3/4x+9/4#

First we can find the point that the tangent line will pass through, which is the point on #f# at #x=2#.

#f(x)=3/2^2=3/4#

The tangent line passes through the point #(2,3/4)#.

Next, we need to find the slope of the tangent line. This is the definition of the derivative, which gives the slope of the tangent line at a given point.

To differentiate the function, we will first need to write the function:

#f(x)=3/x^2=3x^-2#

Then, we will use the power rule to find the derivative. The power rule states that if #f(x)=x^n#, then #f'(x)=nx^(n-1)#.

Constants that the function are multiplied by stay being multiplied by the function, and are not altered. This can be viewed alongside the power rule as if #f(x)=ax^n#, then #f'(x)=anx^(n-1)#.

So, we see that where #f(x)=3x^-2#:

#f'(x)=3(-2)(x^(-2-1))=-6x^-3=-6/x^3#

The slope of the tangent line at #x=2# is:

#f'(2)=-6/2^3=-6/8=-3/4#

So, the tangent line has slope #-3/4# and passes through the point #(2,3/4)#.

A line that passes through the point #(x_1,y_1)# and has slope #m# can be written as:

#y-y_1=m(x-x_1)#

So, with the given information, this becomes:

#y-3/4=-3/4(x-2)#

Simplified, this becomes:

#y=-3/4x+9/4#

We can graph #f# and the tangent line:

graph{(y-3/x^2)(y+3/4x-9/4)=0 [-3.8, 7.296, -1.983, 3.564]}

The line is tangent at #x=2#.

Isaiah Allen · Answer

undefined To find the equation of the line tangent to the function f(x) = 3/x^2 at x = 2, we need to find the derivative of the function and evaluate it at x = 2.

First, let's find the derivative of f(x) = 3/x^2 using the power rule for differentiation.

f'(x) = -6/x^3

Next, we substitute x = 2 into the derivative to find the slope of the tangent line at x = 2.

f'(2) = -6/2^3 = -6/8 = -3/4

So, the slope of the tangent line at x = 2 is -3/4.

Now, we can use the point-slope form of a linear equation to find the equation of the tangent line.

y - y1 = m(x - x1)

Substituting the values, we have:

y - f(2) = (-3/4)(x - 2)

Simplifying further:

y - 3/(2^2) = (-3/4)(x - 2)

y - 3/4 = (-3/4)(x - 2)

y - 3/4 = -3/4x + 3/2

y = -3/4x + 3/2 + 3/4

y = -3/4x + 9/4

Therefore, the equation of the line tangent to f(x) = 3/x^2 at x = 2 is y = -3/4x + 9/4.