Given:
[tex]f(x)=\begin{cases}x^3+4x^2+1,\text{ }ifx\le-4 \\ \sqrt[]{x+5\text{ }},\text{ if }-4a) x=-4.
The condition for the function f(x) continuity at x=a is
[tex]\lim _{x\to a^+}f(x)=\lim _{x\to a^-}f(x)=f(a)[/tex]
Set x=-4, Consider the limit.
[tex]\lim _{x\to-4^+}f(x)=\lim _{x\to-4^+}\sqrt[]{x+5}[/tex]
[tex]\lim _{x\to-4^+}f(x)=\sqrt[]{-4+5}[/tex]
[tex]\lim _{x\to-4^+}f(x)=1[/tex]
Consider the limit.
[tex]\lim _{x\to-4^-}f(x)=\lim _{x\to-4^-}(x^3+4x^2+1)[/tex]
[tex]\lim _{x\to-4^-}f(x)=(-4)^3+4(-4)^2+1[/tex]
[tex]\lim _{x\to-4^-}f(x)=-64+64+1[/tex]
[tex]\lim _{x\to-4^-}f(x)=1[/tex]
Hence we get
[tex]\lim _{x\to-4^-}f(x)=\lim _{x\to-4^+}f(x)=1[/tex]
x= -4 satisfies the continuity condition for the given function.
Hence x= -4 is the continuity point.
b) x=1.
Set x=1, Consider the limit.
[tex]\lim _{x\to1^+}f(x)=\lim _{x\to1^+}\sqrt[]{x+5}[/tex]
[tex]\lim _{x\to1^+}f(x)=\sqrt[]{1+5}[/tex]
[tex]\lim _{x\to1^+}f(x)=\sqrt[]{6}[/tex]
Consider the limit.
[tex]\lim _{x\to1^-}f(x)=\lim _{x\to1^-}\sqrt[]{x+5}[/tex]
[tex]\lim _{x\to1^-}f(x)=\sqrt[]{1+5}[/tex]
[tex]\lim _{x\to1^-}f(x)=\sqrt[]{6}[/tex]
Hence we get
[tex]\lim _{x\to1^-}f(x)=\lim _{x\to1^+}f(x)=\sqrt[]{6}[/tex]
x=1 satisfies the continuity of the given function f(x).
Hence x= 1 is the continuity point.
c) x=3pi/4
Consider the limit.
[tex]\lim _{x\to\frac{3\pi}{4}^+}f(x)=\lim _{x\to\frac{3\pi}{4}^+}\sin x[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^+}f(x)=\sin (\frac{3\pi}{4})[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^+}f(x)=\sin (\pi-\frac{\pi}{4})[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^+}f(x)=\sin (\frac{\pi}{4})[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^+}f(x)=\frac{1}{\sqrt[]{2}}[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^+}f(x)=0.707[/tex]
Consider the limit.
[tex]\lim _{x\to\frac{3\pi}{4}^-}f(x)=\lim _{x\to\frac{3\pi}{4}^-}\sqrt[]{x+5}[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^-}f(x)=\sqrt[]{\frac{3\pi}{4}+5}[/tex][tex]\text{ Use }\pi=180.[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^-}f(x)=\sqrt[]{\frac{3\times180}{4}+5}[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^-}f(x)=\sqrt[]{140}[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^-}f(x)=11.83[/tex]
Hence we get
[tex]0.707\ne11.83[/tex]
[tex]\lim _{x\to\frac{3\pi}{4}^+}f(x)\ne\lim _{x\to\frac{3\pi}{4}^-}f(x)[/tex]
The point x=3pi/4 does not satisfy the continuity of the function.
Hence x=3pi/4 is the discontinuity point of the given function.
The graph of the function is
