For a certain curve $y=f(x)$ satisfying $\frac{d^{2}y}{d{x}^2}=6x-4$, $f(x)$ has a local minimum value $5$ when $x=1$. The global maximum value of $f(x)$ if $0\le x\le 2$, is:

Given $\frac{d^{2}y}{d{x}^2}=6x-4$.Integrate once:
$$\frac{dy}{dx}=3x^2-4x+C_1$$
At $x=1$, there is a local minimum, so $\frac{dy}{dx}=0$:
$$0=3(1{)}^2-4(1)+C_1⟹C_1=1$$
So, $f^{\prime }(x)=3x^2-4x+1=(3x-1)(x-1)$. Critical points are $x=1/3$ and $x=1$.Integrate again:
$$y=x^3-2x^2+x+C_2$$
At $x=1$, $y=5$:
$$5=1-2+1+C_2⟹C_2=5$$
The function is $f(x)=x^3-2x^2+x+5$.To find the global maximum on $[0,2]$, check endpoints and critical points:$f(0)=0-0+0+5=5$$f(2)=2^3-2(2^2)+2+5=8-8+2+5=7$$f(1)=5$ (local min)$f(1/3)=(1/27)-2(1/9)+(1/3)+5=\frac{1-6+9+135}{27}=\frac{139}{27}\approx 5.14$ The maximum value is $7$.