Finding Real Solutions for x-tanh ax = 0 with a>1

[physicsblr]

Can anybody help me with this equation?

I need answer for the question, how many real solutions does the equation x-tanh ax = 0 have for a>1?

 

[tiny-tim]

welcome to pf!

hi physicsblr! welcome to pf!

draw the graph, for 0 ≤ x ≤ 1 (since tanh can't be ≥ 1) …

what is the slope of tanhax ? ​

 

[physicsblr]

Is it a?

 

[physicsblr]

I have a doubt. x=0 is a root of this eq and since its a linear eq, it has only 1 root?

 

[tiny-tim]

what is the slope of tanhax ? ​

Is it a?

no, of course not

how do we find the slope of a graph?​

 

[physicsblr]

oh! slope is zero.

 

[tiny-tim]

how do we find the slope of a graph?

 

[physicsblr]

first derivative of the equation = 0 gives the slope.

 

[tiny-tim]

first derivative of the equation = 0 gives the slope.

well, first derivative of the equation gives the slope

so the slope of tanhax is … ? ​

 

[physicsblr]

1-tanh2 ax; so for that eq. tanh2 ax = 0 is final result.

 

[tiny-tim]

(try using the X² button just above the Reply box )

1-tanh² ax

(= sech²ax)

no

use the chain rule​

 

[physicsblr]

am getting 2 cases. either a=1 or sech2 ax=1.

 

[tiny-tim]

Can anybody help me with this equation?

… how many real solutions does the equation x-tanh ax = 0 have for a>1?

am getting 2 cases. either a=1 or sech2 ax=1.

uhh?

what question are you answering? ​

(and you still need to find the correct derivative of tanh(ax))

 

[physicsblr]

Am unable to get it properly, am no good in mathematics. If u can pls explain it and give an answer since I have an entrance exam next week, I will be very thankful.

 

[tiny-tim]

i] draw the graph of y = x for 0≤x≤1 …

that'll be a square with a diagonal​

ii] find the derivative of y = tanh(ax)

(that is not sech²(ax) … do it again, using the chain rule)

iii] the derivative equals the slope of the graph of y = tanh(ax), so plot the values at x = 0 and 1, and use the slope to find how it starts and finishes …

what do you get? ​

 

[physicsblr]

When x=0, y=1 and there on as x is increasing, value of y is decreasing..

 

[tiny-tim]

When x=0, y=1 and there on as x is increasing, value of y is decreasing..

uhh?

what was unclear about? …​

i] draw the graph of y = x for 0≤x≤1 …

that'll be a square with a diagonal​

ii] find the derivative of y = tanh(ax)

(that is not sech²(ax) … do it again, using the chain rule)

iii] the derivative equals the slope of the graph of y = tanh(ax), so plot the values at x = 0 and 1, and use the slope to find how it starts and finishes …

what do you get? ​