anonymous
  • anonymous
For what value of k is the function f(x)={kx+5, x>4} {x^2 -x, x< or =4} , continuous? JYA
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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agent0smith
  • agent0smith
Find the value of x^2 - x when x=4... 4^2 - 4 = 12 Now find what kx+5 is equal to when x=4... 4k+5 For the function to be continuous: the value of kx+5 (when x=4) must equal the value of x^2 - x (when x=4) So since they're equal, solve this for k: 4k+5 = 12
anonymous
  • anonymous
you're a genius. why didn't i think of that? thank you
agent0smith
  • agent0smith
No prob :)

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anonymous
  • anonymous
so k would equal 7/4? @agent0smith
agent0smith
  • agent0smith
It would.
anonymous
  • anonymous
thank you so much
agent0smith
  • agent0smith
Then you can graph them to see that they're continuous: https://www.google.com/search?q=x%5E2+-x&oq=x%5E2+-x&aqs=chrome..69i57&sourceid=chrome&ie=UTF-8#q=x%5E2-x%2C+7%2F4x%2B5&safe=off Where x=4, the two lines intersect, so the original function f(x)={kx+5, x>4} {x^2 -x, x< or =4} is continuous
anonymous
  • anonymous
no i just needed to know how to do it because im finishing up a math test remediation lol ive got more problems to do:)
anonymous
  • anonymous
i just wasn't sure how to go about that one

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