What is the equation of the line tangent to # f(x)=1/sqrt(x^2+3x+6) # at # x=-1 #?

Question

What is the equation of the line tangent to #  f(x)=1/sqrt(x^2+3x+6) # at #  x=-1 #?

Evelyn Agard · Accepted Answer

#y= -1/16x + 7/16#

Determine the intersection's y-coordinate first.

#f(-1) = 1/sqrt((-1)^2 + 3(-1) + 6)#

#f(-1) = 1/sqrt(4)#

#f(-1) = 1/2#

Now, set yourself apart.

#f(x) = 1/sqrt(x^2 + 3x + 6)#

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

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

Let #y = u^(-1/2)# and #u = x^2 + 3x+ 6#. Then #dy/(du) = -1/2u^(-3/2)# and #(du)/dx = 2x + 3#.

#dy/dx = dy/(du) xx (du)/dx#

#dy/dx= -1/2u^(-3/2) xx (2x + 3)#

#dy/dx= -(2x + 3)/(2(x^2 + 3x + 6)^(3/2)#

The slope at #x= 1# is given by:

#dy/dx|_(x = -1) = -(2(-1) + 3)/(2(-1^2 + 3(-1) + 6)^(3/2))#

#dy/dx|_(x= -1) = -1/16#

Thus, the tangent line's equation is as follows:

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

#y - 1/2 =-1/16(x - (-1))#

#y - 1/2 = -1/16x - 1/16#

#y = -1/16x + 7/16#

I hope this is useful!

Charles Arocha · Answer

undefined To find the equation of the line tangent to the function f(x) = 1/√(x^2 + 3x + 6) at x = -1, we need to find the derivative of the function and evaluate it at x = -1.

The derivative of f(x) = 1/√(x^2 + 3x + 6) can be found using the chain rule.

The derivative is given by: f'(x) = -(x + 3)/[2(x^2 + 3x + 6)^(3/2)]

Evaluating f'(x) at x = -1, we get: f'(-1) = -(-1 + 3)/[2((-1)^2 + 3(-1) + 6)^(3/2)]

Simplifying further, f'(-1) = 2/[2(4)^(3/2)]

Simplifying the denominator, f'(-1) = 2/[2(8)]

Simplifying further, f'(-1) = 1/8

Therefore, the slope of the tangent line at x = -1 is 1/8.

To find the equation of the line, we use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope.

Using the point (-1, f(-1)) = (-1, 1/√(2)), the equation of the tangent line is: y - 1/√(2) = (1/8)(x + 1)

Simplifying further, the equation of the tangent line is: y = (1/8)x + (1/8√(2)) - (1/√(2))