1. Compute the first derivative:

$f(x)=4x^3-11x^2+8x-5$

So,

$f'(x)=12x^2-22x+8=2(6x^2-11x+4)$

Factorizing,

$6x^2-11x+4=(3x-4)(2x-1)$

Hence,

$f'(x)=2(3x-4)(2x-1)$

2. Find critical points:

$f'(x)=0 \Rightarrow (3x-4)(2x-1)=0$

So,

$x=\frac{4}{3},\quad x=\frac{1}{2}$

3. Determine the sign of $f'(x)$ on intervals:

• For $x<\frac{1}{2}$, take $x=0$: $f'(0)=8>0$ So $f$ is increasing on $(-\infty,\frac{1}{2})$.

• For $\frac{1}{2}<x<\frac{4}{3}$, take $x=1$: $f'(1)=12-22+8=-2<0$ So $f$ is decreasing on $\left(\frac{1}{2},\frac{4}{3}\right)$.

• For $x>\frac{4}{3}$, take $x=2$: $f'(2)=48-44+8=12>0$ So $f$ is increasing on $\left(\frac{4}{3},\infty\right)$.

4. Identify local extrema:

• At $x=\frac{1}{2}$, $f'$ changes from positive to negative, so $f$ has a local maximum there, not a minimum.
• At $x=\frac{4}{3}$, $f'$ changes from negative to positive, so $f$ has a local minimum there.

5. Check each option:

• A: has a local minima at $x=\frac{1}{2}$
  False, it has a local maximum there.

• B: has a local minima at $x=\frac{3}{4}$
  False, $x=\frac{3}{4}$ is not even a critical point.

• C: is increasing in $\left(\frac{1}{2},\frac{3}{4}\right)$
  False, $f'(x)<0$ throughout $\left(\frac{1}{2},\frac{4}{3}\right)$, so it is decreasing there.

• D: is decreasing in $\left(\frac{1}{2},\frac{4}{3}\right)$
  True.

Therefore, the correct option is D.