To make f differentiable everywhere, find an ang b.f(x)={ax3x≤2x2+bx&gt;2&nbsp;

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To make f differentiable everywhere, find an ang b.f(x)={ax3x≤2x2+bx&amp;gt;2&amp;nbsp;

Archie Jones · Accepted Answer

We are givenf(x)={ax3x≤2x2+bx&amp;gt;2&amp;nbsp;We want to ensure that the function can be varied everywhere. Since the two functions that make up this piecewise function are polynomials, they are continuous everywhere, which is necessary for this operation, the next thing we want to do is check for continuity at x=2.f(2)=a(2)3=8aAnd the second function gives us:f(2)=22+b=4+bFrom here we know that for the function to be continuous it must be true that:9a=4+bNext we also need make sure the functions first derivative has the same value when x=2, to do so we first find one:f′(2)=3a(2)2=12aFrom the second:f′(2)=2(2)=4So for these two have the same value:12a=4a=412=13If we plug this in the first equation we found we get:8a=4+b83−4=bb=8−123=−43So our conclusion is for the given function to be differentiable everywhere a and b must take the valuesa=13b=−43

Find a ang b such that f is differentiable everywhere. \[f(x)=\begin{cases}ax^3&x\le2\\x^2+b&x>2\end{cases}\]

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