Find an equation of the tangent line to the graph of the function at the given point

Question

Find an equation of the tangent line to the graph of the function at the given point.

y= 4-x²-ln(1/2x+1), point (0 , 4)

Answer 1

I will assume you mean:

which is equal to:

y = 4 - x² - ln[(x+2)/2] or:

y = 4 - x² - ln(x + 2) + ln(2)

So, using the final function, taking the derivative yields:

y' = -2x - 1/(x + 2)

and y'(0) = -1/2

Therefore, using the point-slope form of a line, y - y₁ = m(x - x₁) where (x₁, y₁) = (0, 4) and m = -1/2 we get the equation of the tangent line as:

y - 4 = -1/2(x - 0) or y = (-1/2)x + 4

If you REALLY meant y = 4 - x² - ln(1/(2x + 1)), I would simplify again:

y = 4 - x² - ln(1) + ln(2x + 1)

y = 4 - x² - 0 + ln(2x + 1)

y = 4 - x² + ln(2x + 1)

Then y' = -2x + 2/(2x + 1) and y'(0) = 2

So, using the point-slope form again, y - 4 = 2(x - 0) or y = 2x + 4

Answer 2

y(0)= 4-0^2 -ln(1/2(0) +1) is undefined, not 4. (0,4) is not on the graph

1/2(0) = 1/0 is undefined, +/- infinity

the natural log is undefined

y is undefined when x= 0

the problem was really

y=4-x^2 -ln(1/(2x+1), then

y(0) = 4-0 -ln(1) = 4

y' =-2x+2/(2x+1)

y'(0)= 0+2/1 =2 = slope of the tangent line

2= (y-4)/(x-0)

y-4 = 2x

y = 2x+4 is the tangent line