+0

# Don't you just hate questions like this?

0
342
6

Good morning,

Would somebode please help me out of my misery?. I have this function: f(x)=b^x-1. I cannot sketch the graph given, but basically it's like this:

We have Horizontal asymptote where Y=-1. Then we have (0,0), (-1,1) and (-2,3). So if the 3 coordinates are joined and the curve extended towards the horizontal asymptote, you basically have the graph.

I have 2 questions on this , PLEASE.

the first is to find b

the second is : Use the graph to solve x if: 2^-x=x+3

To anyone who would be able to help, ..please, and thank you are all I can say...

Guest Nov 16, 2015

#5
+5

Alan,

there must be a misunderstanding due to either mis-interpretation or non-proper explanation both of them on my side, for which I profusely apologise,  UNLESS I am mis-understanding you now...

First, please accept my apology for me getting my quadrants all mixed up.

There is only one equation, the equation is f(x)=(b^x)-1.

Its a curved graph which runs through the coordinates as given previously. So, the upper portion lies in the 2nd quadrant and the lower portion in the 4rth, "covering" the area between y=0 and y=-1.

This is how it is given.

THEN, which I did not include in the original thread is:...a second graph, on its own, a line graph with equation g(x)=mx+c. The line is drawn to have (-2,0) and (0,2) as coordinates. (It's all the same question).

The question which needs to be answered is between these 2 graphs, and I assumed it had to do with the "curved" graph, hence the reason I did not include line graph originally.

So, the question states: Use the graph to solve x if 2^-x = x+3.

Please forgive my stupidity if you already undertood things this way, and have already given the correct answer, otherwise, please forgive me for not explaining properly?

Guest Nov 16, 2015
Sort:

#1
+26328
+5

This might help  (note that (1/2)^x is the same thing as 2^-x):

.

Alan  Nov 16, 2015
#2
+5

Thank you Alan,

Your note helped me a lot....I do appreciate!

Guest Nov 16, 2015
#3
+5

Hi Alan,

Just something please...The graph you have lies in the 1st quadrant, while how it should lie, is the upper part in the 4rth and lower part in the 2nd quadrants.

From this we have to use the graph to determine x if: 2^-x=x+3.

Could you please maybe show how "x" could be made the "subject"..(Is that it?)

Guest Nov 16, 2015
#4
+26328
+5

In terms of quadrants, the graph of b^x - 1 lies in the 2nd (x negative, y positive) and 4th (x positive, y negative) quadrants.  The graph of x + 3 lies in the 1st (x positive, y positive), 2nd and 3rd (x negative, y negative) quadrants.  The intersection of these two functions lies in the 2nd quadrant.

Your original question specified using the graph (of the intersection of the two functions) to find the value of x at the point of intersection.  Since you originally spoke of sketching the curve, I assumed you would not be after a very precise value (you can see from the graph that the intersection occurs close to x = -1.4).

When equating the two functions you can't rearrange the result to isolate x.  You need to solve the resulting equality numerically (to get -1.386...).  (Technically, there is an analytical solution, but it involves a non-elementary function called the LambertW function - not something most of us meet in every day life!)

Alan  Nov 16, 2015
#5
+5

Alan,

there must be a misunderstanding due to either mis-interpretation or non-proper explanation both of them on my side, for which I profusely apologise,  UNLESS I am mis-understanding you now...

First, please accept my apology for me getting my quadrants all mixed up.

There is only one equation, the equation is f(x)=(b^x)-1.

Its a curved graph which runs through the coordinates as given previously. So, the upper portion lies in the 2nd quadrant and the lower portion in the 4rth, "covering" the area between y=0 and y=-1.

This is how it is given.

THEN, which I did not include in the original thread is:...a second graph, on its own, a line graph with equation g(x)=mx+c. The line is drawn to have (-2,0) and (0,2) as coordinates. (It's all the same question).

The question which needs to be answered is between these 2 graphs, and I assumed it had to do with the "curved" graph, hence the reason I did not include line graph originally.

So, the question states: Use the graph to solve x if 2^-x = x+3.

Please forgive my stupidity if you already undertood things this way, and have already given the correct answer, otherwise, please forgive me for not explaining properly?

Guest Nov 16, 2015
#6
+26328
+5

Sorry, I misread the second part of your question.  However, 2^-x is not very different from your original f(x), so if you shift the red graph up by 1 everywhere you should find the intersection.

Better than this is to note that 2^-x = f(x)+1, so if 2^-x = x + 3 then f(x) + 1 = x + 3 or f(x) = x + 2.  Lower the straight (blue) line by 1 everywhere (easier than increasing the red line by 1) to find the intersection point. Looks like the two curves would then intersect at x = -1.

Check:  LHS = 2^(- -1) = 2^(+1) = 2.  and RHS = -1+3 = 2

So this checks out ok!

My earlier comment about not being able to isolate x still stands.

Alan  Nov 16, 2015
edited by Alan  Nov 16, 2015
edited by Alan  Nov 16, 2015
edited by Alan  Nov 16, 2015

### 13 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details