Angle curve and line

[Loki123]

I have no idea how to find intercept points because of ln and I also have no idea how to get the angle...

 

[BigBeachBanana]

I have no idea how to find intercept points because of ln and I also have no idea how to get the angle...
View attachment 32293

[imath]\ln x=-x+1 \implies \ln x +x=1[/imath].
Notice that [imath]\ln x+ x[/imath] is a strictly increasing function for [imath]x>0[/imath], and since [imath]x=1[/imath] is in the domain [imath]\therefore x=1[/imath] is the only solution.

 

[Harry_the_cat]

If you sketch the graphs, you will see that (0, 1)(1,0) is the intersection point. You have actually found that. Not sure how you got (1, 0) (0,1) though.

For the angle, you will need to consider the gradients at the intersection point.

 

[Loki123]

[imath]\ln x=-x+1 \implies \ln x +x=1[/imath].
Notice that [imath]\ln x+ x[/imath] is a strictly increasing function for [imath]x>0[/imath], and since [imath]x=1[/imath] is in the domain [imath]\therefore x=1[/imath] is the only solution.

how do I notice that?

 

[BigBeachBanana]

how do I notice that?

Actually, you should know how to show that the function strictly increasing via derivative.
If f′(x)>0 for all x in the interval, then the function f is strictly increasing.

 

[Steven G]

And you should know that ln(x) and x are increasing function as well as knowing that the sum of increasing functions is an increasing function.

 

[Steven G]

If you sketch the graphs, you will see that (0, 1) is the intersection point. You have actually found that. Not sure how you got (1, 0) though.

ln(0) is not 1

 

[Harry_the_cat]

ln(0) is not 1

Yes my mistake, sorry. I meant to say:

If you sketch the graphs, you will see that (1, 0) is the intersection point. You have actually found that. Not sure how you got (0, 1) though.

Had my points around the wrong way.

 

[Harry_the_cat]

If you sketch the graphs, you will see that (0, 1) is the intersection point. You have actually found that. Not sure how you got (1, 0) though.

For the angle, you will need to consider the gradients at the intersection point.

Fixed this up in post #8.

 

[Steven G]

Yes my mistake, sorry. I meant to say:

If you sketch the graphs, you will see that (1, 0) is the intersection point. You have actually found that. Not sure how you got (0, 1) though.

Had my points around the wrong way.

Corner time! We need to bring this back again.

 

[Loki123]

Corner time! We need to bring this back again.

Is this it?

 

[Steven G]

You failed to define k, k1 and k2!

You need to move away from formulas and think instead. You have two lines, one with a slope of 1 and another with a slope of -1. Draw two intersecting lines with these two slopes and the angle between them will be obvious.

 

[Loki123]

You failed to define k, k1 and k2!

You need to move away from formulas and think instead. You have two lines, one with a slope of 1 and another with a slope of -1. Draw two intersecting lines with these two slopes and the angle between them will be obvious.

Oh, that's 90 degrees. I made a mistake in the formula. I get 2/0.

 

[Steven G]

Using a formula is much more prone to errors than just thinking.
A line of slope 1 intersects a line of slope -1 at a 90 degree angle. I do not need a formula to see that and neither do you!