Please help with trig calculus!!

Question

F(x)= -4cos(x)-3 crosses the y axis at P and the positive side of the x axis for the FIRST time at Q.

 

1.Find the coordinates of P and Q

2. Find the STANDARD form equation of the tangent to f(x) at Q

3.Find the standard form equation of PQ

4.Find the point on f(x) between P and Q where the tangent to f(x) is parallel to PQ.

 

I'm totally lost please help!

Answer 1

Hi Linda,

 

If you take the time to visit an online graphing calculator like https://www.desmos.com/calculator and graph the function it will help you to see what is going on.

 

The function crosses the y-axis when x = 0. So, find F(0) to locate the y-intercept.

F(0) = -4cos(0)-3 = -4(1)-3 = -7 (so P has coordinates 0,-7).

 

The function crosses the x-axis when y=0. The expectation for a trig function is that it likely crosses the x-axis many times which is why the problem statement mentions FIRST time.

 

The function crosses the x-axis at any points where its y-value is 0, i.e. F(x) = 0.

 

That means 0=-4cos(x) -3 if solved for x will reveal those x-coordinates.

 

-4cos(x)=3

cos(x) = -3/4

x=arccos(-3/4)

 

The cos function is negative in quadrants 2 and 3, so we would expect the first time the function crosses the x-axis to be a number between pi/2 and pi.

 

Evaluating arccos(-3/4) gives a value of 2.419 (let's just say 2.4).

 

So, Q(2.4, 0).

 

For part 2 we need to know the slope of the tangent line at Q in order to write the equation of that line.

 

F'(x) = 4sin(x) is a formula for the slope of a tangent line to any point (x,F(x)). 

So, F'(2.4) gives the slope of the tangent line at (2.4,0).

 

Now you can use point-slope formula to find the equation of the tangent at (2.4,0).

 

For part 3 find the slope between PQ and use point-slope to write the equation of the line passing through P and Q.

 

Finally for part 4, this is an application of the Intermediate Value Theorem. There is a point on the curve between 0 and 2.4 that has a tangent line with the same slope as PQ.

 

Set F'(x) = slope of PQ and solve for x.

 

4sin(x) = slope PQ

 

sin(x) = slope PQ divided by 4.

 

x = arcsin of that value.Should get something like .809. Find F(0.809) to get the y-coordinate of the point.