Find the values of m and b that make the following function differentiable.

the piecewise function f of x equals x squared when x is less than or equal to three or mx plus b when x is greater than three

Find the values of m and b that make the following function differentiable the piecewise function f of x equals x squared when x is less than or equal to three class=

Respuesta :

For [tex]f[/tex] to be differentiable, it must be continuous, so we need to have

[tex]\displaystyle\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=f(3)[/tex]

By its definition, [tex]f(3)=3^2=9[/tex]. The one-sided limits are

[tex]\displaystyle\lim_{x\to3^-}f(x)=\lim_{x\to3}x^2=9[/tex]

[tex]\displaystyle\lim_{x\to3^+}f(x)=\lim_{x\to3}mx+b=3m+b[/tex]

so we require [tex]3m+b=9[/tex].

In order for [tex]f[/tex] to be differentiable at [tex]x=3[/tex], we also need to have [tex]f'(3)[/tex] exist, which requires that [tex]f'[/tex] also be continuous at [tex]x=3[/tex]. First, compute the derivatives of all pieces of [tex]f[/tex]:

[tex]f'(x)=\begin{cases}2x&\text{for }x<3\\?&\text{for }x=3\\m&\text{for }x>3\end{cases}[/tex]

[tex]f'[/tex] is continuous at [tex]x=3[/tex] if

[tex]\displaystyle\lim_{x\to3^-}f'(x)=\lim_{x\to3^+}f'(x)=f'(3)[/tex]

The one-side limits are

[tex]\displaystyle\lim_{x\to3^-}f'(x)=\lim_{x\to3}2x=6[/tex]

[tex]\displaystyle\lim_{x\to3^+}f'(x)=\lim_{x\to3}m=m[/tex]

so we need to have [tex]m=6[/tex], and moreover [tex]f[/tex] will be differentiable if we set [tex]f'(3)=6[/tex].

So with [tex]m=6[/tex], we must have [tex]3m+b=9\implies b=-9[/tex].

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