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the function is f(x) = 5x^2 then f'(x) = 10x f'(3) = 30
How did you come up with the function?
I looked at x = 3 f(x) = 45.... so I chose 5x^2 substituted x+ 2.8 and came up with 39.2 also its increasing over the domain made me think about a quadratic...
Is there a way to come up with the function without just looking at it? What if it was a bit harder to come up with?
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you can plot the points (x,f(x)) on the number plane... thats an easy way
Thanks
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