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Find a relationship between a and b so that f is continuous at x = 2.
please please please @jdoe0001
substitute in 2 for x and set ax^2+bx=5x-10 4a+2b=10-10, so 4a+2b=0 Which gives 2b=-4a, or 4a=-2b, and to simplify even further, -2a=b or a=-0.5b
can someone check my work?! MEDAL AND FAN!!
f(x) = 5x - 10 for x >= 2 At x = 2: f(x) = 5*2 - 10 = 10 - 10 = 0 f(x) = ax^2 + bx for <= x < 2 The left side limit of f(x) as x->2- is a(2)^2 + b(2) = 4a + 2b In order for f(x) to be continuous at x = 2, the left and right side limits must be the same. 4a + 2b = 0 2a + b = 0
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oh we got diferent answers...
hmm
where did you get 2a+b from?
4a + 2b = 0 divide by 2: 2a + b = 0
gotchaa
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thanks!
you are welcome.
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