Given:
![f(x)=\begin{cases}x^3+4x^2+1,\text{ }ifx\le-4 \\ \sqrt[]{x+5\text{ }},\text{ if }-4a) x=-4.<p>The condition for the function f(x) continuity at x=a is </p>[tex]\lim _(x\to a^+)f(x)=\lim _(x\to a^-)f(x)=f(a)]()
Set x=-4, Consider the limit.
![\lim _(x\to-4^+)f(x)=\lim _(x\to-4^+)\sqrt[]{x+5}](https://img.qammunity.org/2023/formulas/mathematics/college/qcmwrn5atxq1abjpaql4m3acjgm0jm3arw.png)
![\lim _(x\to-4^+)f(x)=\sqrt[]{-4+5}](https://img.qammunity.org/2023/formulas/mathematics/college/sej10qgynvs16751sjmo15kuoi9o163bhg.png)

Consider the limit.




Hence we get

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.
![\lim _(x\to1^+)f(x)=\lim _(x\to1^+)\sqrt[]{x+5}](https://img.qammunity.org/2023/formulas/mathematics/college/fgumiytiv9oqq47erntcxr0tplgy52zxel.png)
![\lim _(x\to1^+)f(x)=\sqrt[]{1+5}](https://img.qammunity.org/2023/formulas/mathematics/college/h1q0502edq8dwyit288l1kapbbw4pg0rk5.png)
![\lim _(x\to1^+)f(x)=\sqrt[]{6}](https://img.qammunity.org/2023/formulas/mathematics/college/roomfmtsflna3ut48buu04whghaknbruuw.png)
Consider the limit.
![\lim _(x\to1^-)f(x)=\lim _(x\to1^-)\sqrt[]{x+5}](https://img.qammunity.org/2023/formulas/mathematics/college/spixsiak37b8h7sm616ng52q634dm7y6fj.png)
![\lim _(x\to1^-)f(x)=\sqrt[]{1+5}](https://img.qammunity.org/2023/formulas/mathematics/college/ckh5u1xuz89shrw0vjlh3ev34tg0r541a3.png)
![\lim _(x\to1^-)f(x)=\sqrt[]{6}](https://img.qammunity.org/2023/formulas/mathematics/college/fjes3tuu4gmfmqw68l61325dgem4uy99g0.png)
Hence we get
![\lim _(x\to1^-)f(x)=\lim _(x\to1^+)f(x)=\sqrt[]{6}](https://img.qammunity.org/2023/formulas/mathematics/college/fof3jxxr6whd36kz76gykeo3ntd3k4oflh.png)
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.




![\lim _{x\to(3\pi)/(4)^+}f(x)=\frac{1}{\sqrt[]{2}}](https://img.qammunity.org/2023/formulas/mathematics/college/9tkcn7nhf0fm7at38hznlwegwp7k2ktgsg.png)

Consider the limit.
![\lim _{x\to(3\pi)/(4)^-}f(x)=\lim _{x\to(3\pi)/(4)^-}\sqrt[]{x+5}](https://img.qammunity.org/2023/formulas/mathematics/college/oq5ez7o07dxkczvlgerpdm5r3cf8eekket.png)
![\lim _{x\to(3\pi)/(4)^-}f(x)=\sqrt[]{(3\pi)/(4)+5}](https://img.qammunity.org/2023/formulas/mathematics/college/n7b0v0ja980prvhfdylz4g4uxllld7gvyu.png)

![\lim _{x\to(3\pi)/(4)^-}f(x)=\sqrt[]{(3*180)/(4)+5}](https://img.qammunity.org/2023/formulas/mathematics/college/iwqiyvzrd8fu31pbuzeisvaj0ipclo7r5a.png)
![\lim _{x\to(3\pi)/(4)^-}f(x)=\sqrt[]{140}](https://img.qammunity.org/2023/formulas/mathematics/college/w2qp2vp4k35ci89i023nmbzhsca2rciivm.png)

Hence we get


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