Question

Hi I’m not sure how to solve this can you help me? I’m in high school Calculus 1, and this is a homework. Thanks!

Answer 1

Given:

Set x=-4, Consider the limit.

`\lim _(x\to-4^+)f(x)=\lim _(x\to-4^+)\sqrt[]{x+5}`

`\lim _(x\to-4^+)f(x)=\sqrt[]{-4+5}`

`\lim _(x\to-4^+)f(x)=1`

Consider the limit.

`\lim _(x\to-4^-)f(x)=\lim _(x\to-4^-)(x^3+4x^2+1)`

`\lim _(x\to-4^-)f(x)=(-4)^3+4(-4)^2+1`

`\lim _(x\to-4^-)f(x)=-64+64+1`

`\lim _(x\to-4^-)f(x)=1`

Hence we get

`\lim _(x\to-4^-)f(x)=\lim _(x\to-4^+)f(x)=1`

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}`

`\lim _(x\to1^+)f(x)=\sqrt[]{1+5}`

`\lim _(x\to1^+)f(x)=\sqrt[]{6}`

Consider the limit.

`\lim _(x\to1^-)f(x)=\lim _(x\to1^-)\sqrt[]{x+5}`

`\lim _(x\to1^-)f(x)=\sqrt[]{1+5}`

`\lim _(x\to1^-)f(x)=\sqrt[]{6}`

Hence we get

`\lim _(x\to1^-)f(x)=\lim _(x\to1^+)f(x)=\sqrt[]{6}`

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)=\lim _{x\to(3\pi)/(4)^+}\sin x`

`\lim _{x\to(3\pi)/(4)^+}f(x)=\sin ((3\pi)/(4))`

`\lim _{x\to(3\pi)/(4)^+}f(x)=\sin (\pi-(\pi)/(4))`

`\lim _{x\to(3\pi)/(4)^+}f(x)=\sin ((\pi)/(4))`

`\lim _{x\to(3\pi)/(4)^+}f(x)=\frac{1}{\sqrt[]{2}}`

`\lim _{x\to(3\pi)/(4)^+}f(x)=0.707`

Consider the limit.

`\lim _{x\to(3\pi)/(4)^-}f(x)=\lim _{x\to(3\pi)/(4)^-}\sqrt[]{x+5}`

`\lim _{x\to(3\pi)/(4)^-}f(x)=\sqrt[]{(3\pi)/(4)+5}`
`\text{ Use }\pi=180.`

`\lim _{x\to(3\pi)/(4)^-}f(x)=\sqrt[]{(3*180)/(4)+5}`

`\lim _{x\to(3\pi)/(4)^-}f(x)=\sqrt[]{140}`

`\lim _{x\to(3\pi)/(4)^-}f(x)=11.83`

Hence we get

`0.707\\e11.83`

`\lim _{x\to(3\pi)/(4)^+}f(x)\\e\lim _{x\to(3\pi)/(4)^-}f(x)`

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