a = 4 12 = 1 3. If we plug this in the first equation we found we get: 8 a = 4 + b. 8 3 − 4 = b. b = 8 − 12 3 = − 4 3. So our conclusion is for the given function to be differentiable everywhere a and b

We will solve first for a by making it differentiable; then we can solve for b by making it continuous. This is a very standard Calc AB question, by the way. The function is

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SOLUTION: Find a and b so that the function f (x)= {3x^3−8x^2+8;x is both continuous and differentiable. a= b=. is both continuous and differentiable. You can put this

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find the constant m and b in the linear function f(x)=mx+b so that f(6)=9 and the straight line represents by f has slope -4. View Answer. Find a and b such that f is differentiable everywhere.