What is the tangent line of y = cos x at x = p/4?

Question

1. What is the tangent line of y = cos x at x = π/4?

2. What is the derivative of (5x + 2)(x³ + 4x)?

3. Construct an f(x) such that lim f(x) = ∞ but lim (x-3)f(x) = 0

                                                                     x→3      x→3

Answer 1

1)

 

First, we need to evaluate y when x=π/4.  This will give us the target coordinate point.

 

The coordinate point we need is (π / 4 , √(2) / 2)

 

Next, we find the derivative of the function y.

 

y ' = -sin(x)

 

 

Plug in the value of x into the derivative.  This will be the slope of the tangent line.

 

y' = -sin(π/4)

y' = - √(2) / 2

 

Now, we plug in the slope and coordinate point into the slope intercept form of the equation to find b.

 

y = mx + b

 

√(2) / 2 = (-√(2) / 2)(π / 4) + b

 

√(2) / 2 = (-π√(2) / 8) + b

 

(√(2) / 2) + (π√(2) / 8) = b

 

[4√(2) + π√(2)] / 8 = b

 

[(4 + π)√(2)] / 8 = b

 

 

The tangent line is

 

y = (-√(2) / 2)x + [(4 + π)√(2) / 8]

 

 

Feel free to simplify this tangent line.

 

 

2)

 

We need to use the product rule because we have two factors that are being multiplied together, right?  Well not really.  We can use FOIL to expand the expression, then derive term by term using the power rule.

 

d/dx[(5x + 2)(x³ + 4x)] =

 

d/dx (5x⁴ + 2x³ + 20x² + 8x) =

 

20x³ + 6x² + 40x + 8

 

 

3)

 

When finding the limit of a function, we are looking for where the function has an horizontal asymptote.  If the limit is infinity, then there is no horizontal asymptote because the horizontal is the limit where the function can reach.  For example, if the limit of f(x) is 3, then f(x) can reach no higher than 3, but never touches 3.

 

Recall that in pre-calculus that if the numerator part of a rational function has a degree higher than the denominator's, then there is no horizontal asymptote.  The limit does not exist.  If the numerator part of a rational function has a degree lower that the denominator's, then the horizontal asymptote is zero.  The limit is then zero.

 

According to the problem, as x approaches 3, the limit of (x - 3)f(x) is zero.  So if we plug in 3, we get 0f(x), which is 0.

 

 

f(x) = (x - 3)(x + 3) / x

 

f(x) = (x² - 9) / x

Answer 2

Question 1) We need to understand what the tangent *is*. The tangent at a point of any function is its slope at that point. This should ring a bell that we need to find the derivative.

 

d/dx [cos (x)] = -sin(x)

 

Now evaluate at π/4 and we find the the derivative (slope) is -sqrt(2)/2.

 

We wanted an equation for the line, so we should think of using the point-slope format, which means we need the value of cos(x) at π/4. It turns out that this is also sqrt(2)/2 because that angle is 45 degrees and sin and cos are the same at that angle.

 

Now our equation for the line is

 

y - sqrt(2)/2 = - sqrt(2)/2 * (x - π/4)

 

Question 2) asks for the derivative of a complicated function. We should use the chain and product rules for this

 

d/dx [ (5x + 2)(x³ + 4x) ] =

 

(5x + 2) d/dx (x3 + 4x) + d/dx(5x + 2) * (x3 + 4x) 

 

(5x+2) * (3x² + 4) + (5) * (x³ +4x)

 

You can simplify this on your own 

 

Question 3) Is unclear from the way you typed it in. Can you please try using words to describe the situation? Does it have something to do with the limit from the left and the limit from the right as they approach 3? I'm not sure what you're asking for.

 

I hope this helps!