find all points on the graph of f at which the tangent line is horizontal
a) f(x)=(x+1)/(x^2+3)
b) y=cos(x)+sin(x),0≤x≤2π
For the tangent line to be horizontal, the slope must be zero
f(x)=(x+1)/(x^2+3)
f'(x) = ( (x^2 + 3)(1) - (x+1)(2x) )/(x^2 + 3)^2
= 0
x^2 + 3 - 2x^2 - 2x = 0
-x^2 - 2x + 3 = 0
x^2 + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 , x = 1
if x = -3, f(x) = (-3+1)/(9 + 3) = -2/12 = -1/6
so one point is (-3,-1/6)
if x = 1 ......... (your turn)
y = cosx + sinx
dy/dx = -sinx + cosx = 0 for horizonatal tangent
sinx = cosx
sinx/cosx = 1
tanx = 1
x = π/4, 5π/4, (45°, 225°) ,
(9π/4 would be the next one, but that's outside domain)
if x = π/4
y = sinπ/4 + cosπ/4 = √2/2 + √2/2 = √2
so one point is (π/4, √2)
your turn to find the other point
or, note that cosx + sinx = √2 sin(x + π/4)
so the tangent is horizontal when
cos(x + π/4) = 0
That is when x + π/4 = π/2 or 3π/2
one solution, given above, is x = π/4
your turn to find the other
To find the points on the graph of a function where the tangent line is horizontal, we need to find the values of x where the derivative of the function is equal to zero.
a) Let's find the derivative of f(x) = (x+1)/(x^2+3) to determine the slope of the tangent line at any point on the graph:
First, rewrite the function as f(x) = (x+1)*(x^2+3)^(-1).
Then, apply the quotient rule to differentiate the function. The quotient rule states if we have two functions, u(x) and v(x), then the derivative of u(x) divided by v(x) is given by [u'(x) * v(x) - v'(x) * u(x)] / [v(x)]^2.
For f(x) = (x+1)*(x^2+3)^(-1), applying the quotient rule:
f'(x) = [(1) * (x^2+3) - (x+1) * (2x)] / [(x^2+3)^2]
= (x^2 + 3 - 2x^2 - 2) / (x^2+3)^2
= (-x^2 - 2x + 1) / (x^2+3)^2
To find the points where the tangent line is horizontal, we set f'(x) = 0 and solve for x:
(-x^2 - 2x + 1) / (x^2+3)^2 = 0
Multiplying both sides by (x^2+3)^2, we get:
-x^2 - 2x + 1 = 0
Rearranging the equation, we have:
x^2 + 2x - 1 = 0
We can solve this quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 2, and c = -1. Plugging these values into the quadratic formula:
x = (-2 ± √(2^2 - 4(1)(-1))) / (2 * 1)
x = (-2 ± √(4 + 4)) / 2
x = (-2 ± √8) / 2
x = (-2 ± 2√2) / 2
x = -1 ± √2
So, the two values of x where the tangent line is horizontal for the function f(x) = (x+1)/(x^2+3) are: x = -1 + √2 and x = -1 - √2.
b) To find the points on the graph of y = cos(x) + sin(x) where the tangent line is horizontal, we need to find the values of x where the derivative of the function dy/dx is equal to zero.
First, let's find the derivative dy/dx of y = cos(x) + sin(x). The derivative of sin(x) is cos(x), and the derivative of cos(x) is -sin(x). So the derivative of y with respect to x is:
dy/dx = -sin(x) + cos(x)
Now, to find the points where the tangent line is horizontal, we set dy/dx = 0 and solve for x:
-sin(x) + cos(x) = 0
Rearranging the equation, we have:
sin(x) = cos(x)
To solve this equation, we can use the trigonometric identity tan(x) = sin(x)/cos(x).
Dividing both sides of the equation by cos(x), we get:
tan(x) = 1
Now, we know that tan(x) = 1 when x = π/4.
However, we need to consider the given range 0≤x≤2π. This means that the points where the tangent line is horizontal occurs at x = π/4 within the specified range.
Therefore, the point on the graph of y = cos(x) + sin(x) where the tangent line is horizontal is (π/4, cos(π/4) + sin(π/4)).