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Find the value of constant K that makes the function below continuous at x = 5?? (x^2-5x)/(x-5) x<5 f(x) kx-3 x>equal to 5
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This is a lot like the other question. For f(x) to be continuous at x=5, all function parts around there must be equal at x=5. So to the left hand side of x, you have: \[f(x) = \frac{x^2-5x}{x-5}\]You need to solve for x=5. (tip: try factoring the numerator) The next step is to say that all parts of the function have to be equal: \[\frac{x^2-5x}{x-5}=kx-3 \text{ ; x=5}\]
ok so when i factor the equation it comes to just being x. is that right?
It is indeed!
yay i got it! 8/5. thank youu
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