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Margaret [11]
3 years ago
9

Find the values of m and b that make the following function differentiable. the piecewise function f of x equals x cubed when x

is less than or equal to one or mx plus b when x is greater than one.
Mathematics
1 answer:
aalyn [17]3 years ago
5 0

f(x)=\begin{cases}x^3&\text{for }x\le1\\mx+b&\text{for }x>1\end{cases}

f must be continuous in order to be differentiable, so we need to have

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1^+}f(x)=f(1)

By its definition, f(1)=1^3=1, and

\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}x^3=1

\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(mx+b)=m+b

so that \boxed{m+b=1}.

We want the derivative to exist at x=1, which requires that we pick an appropriate value for f'(1) so that f'(x) is also continuous. At the moment, we know

f'(x)=\begin{cases}3x^2&\text{for }x1\end{cases}{/tex]so we need to pick [tex]f'(1) such that

\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1^+}f'(x)

We have

\displaystyle\lim_{x\to1^-}f'(x)=\lim_{x\to1}3x^2=3

\displaystyle\lim_{x\to1^+}f'(x)=\lim_{x\to1}m=m

so that \boxed{m=3} (which means we need to pick f'(1)=3) and so m+b=1\implies\boxed{b=-2}

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