Question

I want to know if my answer for this problem is correct. I chose D.

Answer 1

You need to post the problem

Answer 2

let me put the image.. can you see it?

Answer 3

No not d, you need an a so that both equations yield the same result at x=2.

Answer 4

@stephy321
What you'd like to do in order to assure continuity at x=2 is to evaluate both functions of the piecewise function, and equate the results (in terms of a).
Solve the resulting equation to find a.
Example:
Find a to make f(x) a piecewise continuous function:
f(x)=x^2 when x<=3, and f(x)=x+a when x>3
step 1:
Identify the possible discontinuity according to the given limits: x=3.
step 2:
evaluate f(x) of both functions and equate them.
f(3)=x^2=3^2
f(3)=x+a=3+a
equate: 3^2=3+a
step 3: solve
=> 3^2=3+a => 3+a=9 => a=9-3=6
Answer: with a=6, f(x) is a piecewise continuous function.

Answer 5

\[L.H.L=\lim_{x \rightarrow 2-}(x^3+a^3)\]
put x=2-h,h>0 as \[x \rightarrow 2-\]
\[L.H.L=\lim_{h \rightarrow 0}\left\{ \left( 2-h \right)^3+a^3 \right\}=8+a^3\]
\[R.H.L=\lim_{x \rightarrow 2+}\left( 2-2x \right)\]
put x=2+h,
\[h \rightarrow0~as~x \rightarrow2+\]

Answer 6

\[R.H.L.=\lim_{h \rightarrow 0}\left\{ 4-2\left( 2+h \right) \right\}=\lim_{h \rightarrow 0}\left( 4-4-2h \right)=0\]
as f(x) is continuous so L.H.L=R.H.L
\[8+a^3=0,a^3=-8=\left( -2 \right)^3,a=-2\]

Answer 7

write h(x) in place of f(x)