let (f) be defined by (f(x)=\begin{cases}3x^{3}-2m, & xleq - 1\\3x^{2}+5m, & x > - 1end{cases}) (a) find (in…

let (f) be defined by (f(x)=\begin{cases}3x^{3}-2m, & xleq - 1\\3x^{2}+5m, & x > - 1end{cases}) (a) find (in terms of (m)) (lim_{x\rightarrow - 1^{+}}f(x)). limit = (b) find (in terms of (m)) (lim_{x\rightarrow - 1^{-}}f(x)). limit = (c) find the value of (m) so that (lim_{x\rightarrow - 1^{-}}f(x)=lim_{x\rightarrow - 1^{+}}f(x)). (m =) (that is: find the value of (m) that makes the function (f(x)) continuous.) note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 2 times. your overall recorded score is 33%. you have unlimited attempts remaining.

let (f) be defined by (f(x)=\begin{cases}3x^{3}-2m, & xleq - 1\\3x^{2}+5m, & x > - 1end{cases}) (a) find (in terms of (m)) (lim_{x\rightarrow - 1^{+}}f(x)). limit = (b) find (in terms of (m)) (lim_{x\rightarrow - 1^{-}}f(x)). limit = (c) find the value of (m) so that (lim_{x\rightarrow - 1^{-}}f(x)=lim_{x\rightarrow - 1^{+}}f(x)). (m =) (that is: find the value of (m) that makes the function (f(x)) continuous.) note: you can earn partial credit on this problem. preview my answers submit answers you have attempted this problem 2 times. your overall recorded score is 33%. you have unlimited attempts remaining.

Answer

Explanation:

Step1: Find right - hand limit

For $\lim_{x\rightarrow - 1^{+}}f(x)$, since $x>-1$, we use $f(x)=3x^{2}+5m$. Substitute $x = - 1$ into $3x^{2}+5m$: $3(-1)^{2}+5m=3 + 5m$.

Step2: Find left - hand limit

For $\lim_{x\rightarrow - 1^{-}}f(x)$, since $x\leq - 1$, we use $f(x)=3x^{3}-2m$. Substitute $x=-1$ into $3x^{3}-2m$: $3(-1)^{3}-2m=-3 - 2m$.

Step3: Set left - hand and right - hand limits equal

Set $-3 - 2m=3 + 5m$. Add $2m$ to both sides: $-3=3 + 7m$. Subtract 3 from both sides: $-6 = 7m$. Solve for $m$: $m=-\frac{6}{7}$.

Answer:

$m =-\frac{6}{7}$