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First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and $$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$ Since $g(t)$ is increasing in $t\ge1$, wlog $t\le1$.

Next, change the variables and the constants according to the formulas $$u=(t-1)^2,\quad A:=a^2,\quad c_2:=\min(1,c^2),$$ so that $u\in[0,1]$ and $t=1-\sqrt u$. So, the problem reduces to minimizing $$h(u):=\sqrt{u + A} + b(1-\sqrt u)$$ in $u\in[0,c_2]$ given $A\ge0$, $b\ge0$, and $c_2\in[0,1]$.

Note that $h'(u)$ has the same sign for real $u>0$ as $\frac u{A+u}-b^2$, which increases from $-b^2$ to $1-b^2$ as $u$ increases from $0$ to $\infty$. So, the only critical point of $h$ on $(0,\infty)$ is $$u_*:=A\frac{b^2}{1-b^2}$$ if $0\le b<1$, and no critical points of $h$ on $(0,\infty)$ if $b\ge1$.

Letting now $h_{\min}$ denote the minimum of $h$ on $[0,c_2]$, we see that the desired minimum of $g$ is $$ h_{\min}= \begin{cases} h(u_*)=b+\sqrt{(1-b^2)A}&\text{ if }0\le b<1\ \&\ u_*\le c_2, \\ h(c_2)=\sqrt{c_2 + A} + b(1-\sqrt{c_2})&\text{ if }b\ge1\ \text{or}\ (0\le b<1\ \&\ u_*>c_2). \end{cases} $$ This can be simplified: $$ h_{\min}= \begin{cases} h(u_*)=b+a\sqrt{1-b^2}&\text{ if }0\le b\sqrt{\dfrac{c_2}{a^2+c_2}, \\ h(c_2)=\sqrt{c_2 + a^2} + b(1-\sqrt{c_2})&\text{ otherwise }. \end{cases} $$

First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and $$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$ Since $g(t)$ is increasing in $t\ge1$, wlog $t\le1$.

Next, change the variables and the constants according to the formulas $$u=(t-1)^2,\quad A:=a^2,\quad c_2:=\min(1,c^2),$$ so that $u\in[0,1]$ and $t=1-\sqrt u$. So, the problem reduces to minimizing $$h(u):=\sqrt{u + A} + b(1-\sqrt u)$$ in $u\in[0,c_2]$ given $A\ge0$, $b\ge0$, and $c_2\in[0,1]$.

Note that $h'(u)$ has the same sign for real $u>0$ as $\frac u{A+u}-b^2$, which increases from $-b^2$ to $1-b^2$ as $u$ increases from $0$ to $\infty$. So, the only critical point of $h$ on $(0,\infty)$ is $$u_*:=A\frac{b^2}{1-b^2}$$ if $0\le b<1$, and no critical points of $h$ on $(0,\infty)$ if $b\ge1$.

Letting now $h_{\min}$ denote the minimum of $h$ on $[0,c_2]$, we see that the desired minimum of $g$ is $$ h_{\min}= \begin{cases} h(u_*)=b+\sqrt{(1-b^2)A}&\text{ if }0\le b<1\ \&\ u_*\le c_2, \\ h(c_2)=\sqrt{c_2 + A} + b(1-\sqrt{c_2})&\text{ if }b\ge1\ \text{or}\ (0\le b<1\ \&\ u_*>c_2). \end{cases} $$ This can be simplified: $$ h_{\min}= \begin{cases} b+a\sqrt{1-b^2}&\text{ if }b^2\le\dfrac{c_2}{a^2+c_2}, \\ \sqrt{c_2 + a^2} + b(1-\sqrt{c_2})&\text{ otherwise}. \end{cases} $$

First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and $$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$ Since $g(t)$ is increasing in $t\ge1$, wlog $t\le1$.

Next, change the variables and the constants according to the formulas $$u=(t-1)^2,\quad A:=a^2,\quad c_2:=\min(1,c^2),$$ so that $u\in[0,1]$ and $t=1-\sqrt u$. So, the problem reduces to minimizing $$h(u):=\sqrt{u + A} + b(1-\sqrt u)$$ in $u\in[0,c_2]$ given $A\ge0$, $b\ge0$, and $c_2\in[0,1]$.

Note that $h'(u)$ has the same sign for real $u>0$ as $\frac u{A+u}-b^2$, which increases from $-b^2$ to $1-b^2$ as $u$ increases from $0$ to $\infty$. So, the only critical point of $h$ on $(0,\infty)$ is $$u_*:=A\frac{b^2}{1-b^2}$$ if $0\le b<1$, and no critical points of $h$ on $(0,\infty)$ if $b\ge1$.

Letting now $h_{\min}$ denote the minimum of $h$ on $[0,c_2]$, we see that the desired minimum of $g$ is $$ h_{\min}= \begin{cases} h(u_*)=b+\sqrt{(1-b^2)A}&\text{ if }0\le b<1\ \&\ u_*\le c_2, \\ h(c_2)=\sqrt{c_2 + A} + b(1-\sqrt{c_2})&\text{ if }b\ge1\ \text{or}\ (0\le b<1\ \&\ u_*>c_2). \end{cases} $$

First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and $$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$ Since $g(t)$ is increasing in $t\ge1$, wlog $t\le1$.

Next, change the variables and the constants according to the formulas $$u=(t-1)^2,\quad A:=a^2,\quad c_2:=\min(1,c^2),$$ so that $u\in[0,1]$ and $t=1-\sqrt u$. So, the problem reduces to minimizing $$h(u):=\sqrt{u + A} + b(1-\sqrt u)$$ in $u\in[0,c_2]$ given $A\ge0$, $b\ge0$, and $c_2\in[0,1]$.

Note that $h'(u)$ has the same sign for real $u>0$ as $\frac u{A+u}-b^2$, which increases from $-b^2$ to $1-b^2$ as $u$ increases from $0$ to $\infty$. So, the only critical point of $h$ on $(0,\infty)$ is $$u_*:=A\frac{b^2}{1-b^2}$$ if $0\le b<1$, and no critical points of $h$ on $(0,\infty)$ if $b\ge1$.

Letting now $h_{\min}$ denote the minimum of $h$ on $[0,c_2]$, we see that the desired minimum of $g$ is $$ h_{\min}= \begin{cases} h(u_*)=b+\sqrt{(1-b^2)A}&\text{ if }0\le b<1\ \&\ u_*\le c_2, \\ h(c_2)=\sqrt{c_2 + A} + b(1-\sqrt{c_2})&\text{ if }b\ge1\ \text{or}\ (0\le b<1\ \&\ u_*>c_2). \end{cases} $$ This can be simplified: $$ h_{\min}= \begin{cases} h(u_*)=b+a\sqrt{1-b^2}&\text{ if }0\le b\sqrt{\dfrac{c_2}{a^2+c_2}, \\ h(c_2)=\sqrt{c_2 + a^2} + b(1-\sqrt{c_2})&\text{ otherwise }. \end{cases} $$

First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and $$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$ Since $g(t)$ is increasing in $t\ge1$, wlog $t\le1$.

Next, change the variables and the constants according to the formulas $$u=(t-1)^2,\quad A:=a^2,\quad c_2:=\min(1,c^2),$$ so that $u\in[0,1]$ and $t=1-\sqrt u$. So, the problem reduces to minimizing $$h(u):=\sqrt{u + A} + b(1-\sqrt u)$$ in $u\in[0,c_2]$ given $A\ge0$, $b\ge0$, and $c_2\in[0,1]$.

Note that $h'(u)$ has the same sign for real $u>0$ as $\frac u{A+u}-b^2$, which increases from $-b^2$ to $1-b^2$ as $u$ increases from $0$ to $\infty$. So, the only critical point of $h$ on $(0,\infty)$ is $$u_*:=A\frac{b^2}{1-b^2}$$ if $b<1$, and no critical points of $h$ on $(0,\infty)$ if $b\ge1$.

Letting now $h_{\min}$ denote the minimum of $h$ on $[0,c_2]$, we see that the desired minimum of $g$ is $$ h_{\min}= \begin{cases} h(u_*)=b+\sqrt{(1-b^2)A}&\text{ if }0\le b<1\ \&\ u_*\le c_2, \\ h(c_2)=\sqrt{c_2 + A} + b(1-\sqrt{c_2})&\text{ if }b\ge1\ \text{or}\ (0\le b<1\ \&\ u_*>c_2). \end{cases} $$

First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and $$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$ Since $g(t)$ is increasing in $t\ge1$, wlog $t\le1$.

Next, change the variables and the constants according to the formulas $$u=(t-1)^2,\quad A:=a^2,\quad c_2:=\min(1,c^2),$$ so that $u\in[0,1]$ and $t=1-\sqrt u$. So, the problem reduces to minimizing $$h(u):=\sqrt{u + A} + b(1-\sqrt u)$$ in $u\in[0,c_2]$ given $A\ge0$, $b\ge0$, and $c_2\in[0,1]$.

Note that $h'(u)$ has the same sign for real $u>0$ as $\frac u{A+u}-b^2$, which increases from $-b^2$ to $1-b^2$ as $u$ increases from $0$ to $\infty$. So, the only critical point of $h$ on $(0,\infty)$ is $$u_*:=A\frac{b^2}{1-b^2}$$ if $0\le b<1$, and no critical points of $h$ on $(0,\infty)$ if $b\ge1$.

Letting now $h_{\min}$ denote the minimum of $h$ on $[0,c_2]$, we see that the desired minimum of $g$ is $$ h_{\min}= \begin{cases} h(u_*)=b+\sqrt{(1-b^2)A}&\text{ if }0\le b<1\ \&\ u_*\le c_2, \\ h(c_2)=\sqrt{c_2 + A} + b(1-\sqrt{c_2})&\text{ if }b\ge1\ \text{or}\ (0\le b<1\ \&\ u_*>c_2). \end{cases} $$