A line perpendicular to L0: y=-x/2+1 is L1: y=2x+k, i.e. the slope is 2.
In fact, lines perpendicular to any line y=mx+b is y=-x/m + k, where k is a constant.
To find the tangents to g(x):
1. find the derivative g'(x).
2. equate g'(x) with the slope of L1 (i.e. 2).
3. Solve for the value(s) of x which satisfy g'(x)=2, say x0, x1, ...
4. Equate g(x0)=2x+k to find k, and hence the equation of the tangent L1.
5. Repeat for other values x1, x2...
write the equation (in slope intercept form) of the line perpendiculaer to
y=-X/2+1 and tangent to g(x)=3X^5+2X-1
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