We have, ‌f(x)=(2x−1)(3x+2)(4x−3) ‌f′(x)=2(3x+2)(4x−3) ‌+3(2x−1)(4x−3) ‌+4(2x−1)(3x+2) ‌=2[12x2−x−6]+3[8x2−10x+3] ‌‌‌+4[6x2+x−2] ‌=24x2−2x−12+24x2−30x+9 ‌+ ‌+24x2+4x−8 ‌f′(x)=72x2−28x−11 For Rolle's theorem, c is such that f′(c)=0,‌ where ‌c∈[‌