Given:
[tex]f(x)=\begin{cases}-5x^2+6x\ldots\ldots x\leq0 \\ 4\ln (1-4x)+5\ldots.\text{ x>0}\end{cases}[/tex]
Sol:
Function for limit :
[tex]0^-[/tex][tex]f(x)=-5x^2+6x[/tex]
Derivatives of function is:
[tex]\begin{gathered} f^{}(x)=-5x^2+6x \\ f^{\prime}(x)=-10x+6 \end{gathered}[/tex][tex]\begin{gathered} \lim _{x\to0^-}f^{\prime}(x)=\lim _{x\to0^-}(-10x+6) \\ =6 \end{gathered}[/tex]
Function is:
[tex]\begin{gathered} f(x)=4\ln (1-4x)+5 \\ \text{for }0^+^{} \end{gathered}[/tex][tex]\begin{gathered} f(x)=4\ln (1-4x)+5 \\ f^{\prime}(x)=\frac{4}{1-4x}(-4) \\ f^{\prime}(x)=-\frac{16}{1-4x} \end{gathered}[/tex][tex]\begin{gathered} \lim _{x\to0^+}f^{\prime}(x)=\lim _{x\to0^+}(-\frac{16}{1-4x}) \\ =-16 \end{gathered}[/tex]
