The cross-section of a precision-engineered lens is modeled by a curve y=f(x)y = f(x)y=f(x), for x>0x > 0x>0. The rate of change of the gradient of the profile is given by
f′′(x)=154x7−6x f''(x) = \frac{15}{4\sqrt{x^7}} - 6x f′′(x)=4x715−6xA point P(1,1.5)P(1, 1.5)P(1,1.5) lies on the boundary of the lens profile.
Given that the gradient of the curve f′(x)=0.75f'(x) = 0.75f′(x)=0.75 at point PPP,
find the equation of the normal at PPP, writing your answer in the form y=mx+cy = mx + cy=mx+c, where mmm and ccc are constants,
find f(x)f(x)f(x).
