Skip to main content
Mathematica

Return to Question

edited tags
Link
J. M.'s missing motivation ♦
J. M.'s missing motivation ♦
  • 125.5k
  • 11
  • 407
  • 581
deleted 10 characters in body
Source Link
xyz
xyz
  • 655
  • 4
  • 38
  • 117

I need a help to find k here:

f(x)=-k^x + x^2 It is not too simple. I have interest to find the single value of "k" where, inside interval {x,2,3}, the function have only one value to x when f(x) = 0. In fact, using try and error I have already found the value: ratio = -2.0870652286345332^x + x^2 Plot[ratio, {x, 2.715, 2.720}] FindMaximum[ratio, {x, 2.715, 2.720}] NSolve[ratio == 0 && x >= 2, x] ![XK Graph][1] I need to know, a better way to reach the value -2.0870652286345332 to k. Optimization? Solving? I was unable to got it. Any insight are welcome. Best regards [1]: https://i.sstatic.net/AZJtG.jpg

$f(x)=-k^x + x^2$

It is not too simple. I have interest to find the single value of "k" where, inside interval {x,2,3}, the function have only one value to x when f(x) = 0.

In fact, using try and error I have already found the value:

ratio = -2.0870652286345332^x + x^2

Plot[ratio, {x, 2.715, 2.720}] 

FindMaximum[ratio, {x, 2.715, 2.720}]

NSolve[ratio == 0 && x >= 2, x]

XK Graph

I need to know, a better way to reach the value -2.0870652286345332 to k. Optimization? Solving? I was unable to got it.

Any insight are welcome.

Best regards

I need a help to find k here:

f(x)=-k^x + x^2 It is not too simple. I have interest to find the single value of "k" where, inside interval {x,2,3}, the function have only one value to x when f(x) = 0. In fact, using try and error I have already found the value: ratio = -2.0870652286345332^x + x^2 Plot[ratio, {x, 2.715, 2.720}] FindMaximum[ratio, {x, 2.715, 2.720}] NSolve[ratio == 0 && x >= 2, x] ![XK Graph][1] I need to know, a better way to reach the value -2.0870652286345332 to k. Optimization? Solving? I was unable to got it. Any insight are welcome. Best regards [1]: https://i.sstatic.net/AZJtG.jpg

I need a help to find k here:

$f(x)=-k^x + x^2$

It is not too simple. I have interest to find the single value of "k" where, inside interval {x,2,3}, the function have only one value to x when f(x) = 0.

In fact, using try and error I have already found the value:

ratio = -2.0870652286345332^x + x^2

Plot[ratio, {x, 2.715, 2.720}] 

FindMaximum[ratio, {x, 2.715, 2.720}]

NSolve[ratio == 0 && x >= 2, x]

XK Graph

I need to know, a better way to reach the value -2.0870652286345332 to k. Optimization? Solving? I was unable to got it.

Any insight are welcome.

Best regards

Source Link
Rodolfo Silva
Rodolfo Silva
  • 21
  • 1

Only one root at specific interval

I need a help to find k here:

f(x)=-k^x + x^2 It is not too simple. I have interest to find the single value of "k" where, inside interval {x,2,3}, the function have only one value to x when f(x) = 0. In fact, using try and error I have already found the value: ratio = -2.0870652286345332^x + x^2 Plot[ratio, {x, 2.715, 2.720}] FindMaximum[ratio, {x, 2.715, 2.720}] NSolve[ratio == 0 && x >= 2, x] ![XK Graph][1] I need to know, a better way to reach the value -2.0870652286345332 to k. Optimization? Solving? I was unable to got it. Any insight are welcome. Best regards [1]: https://i.sstatic.net/AZJtG.jpg