Let `-1lt=plt=1` . Show that the equation `4x^3-3x-p=0` has a unique root in the interval [1/2,1] and identify it.

Explanation

We have `4x^(3)-3x-p=0`
`implies4x^(3)-3x=p`
Let `f(x)=4x^(3)-3x` and `g(x)=p`
`:'f'(x)=0`
`implies12x^(2)-3=0`
`impliesx=-1/2, -1/2` and `f''(x)=24x`
`:.f''(-1/2)=-12lt0` and `f''(1/2)=12gt0`
`:.f(x)` has local maximum at `(x=-1/2)` and local minimum at `(x=1/2)`
Also `f(-1/2)=-4/8+3/2=1` and `f(1/2)=4/8-3/2=-1`
Clearly `phi' (x)gt0` for `x epsilon[1/21]`
Hence `phi(x)` can have atmost one root in `[1/2,1]`
Also `phi(1/2)=-1-p` and `phi(1)=1-p`
`:.phi(1/2)phi(1)=-(1p-^(2))=(p^(2)-1)le0 [ :' -1leple1]`
Since `phi(x)` being a polynomial, continuous on `[1/2,1]` and `phi(1/2)phi(1)le0`. Therefore, by intermediate value theorem `phi(x)` has atleast one root i `[1/2,1]`
Hence `phi(x)` has exactly one root in `[1/2,1]`