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

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...

To find the equation of the line that is perpendicular to y = -x/2 + 1 and tangent to g(x) = 3x^5 + 2x - 1, we need to find the slope and the point of tangency.

First, let's determine the slope of the line y = -x/2 + 1. The slope-intercept form of a line is y = mx + b, where m represents the slope. In this case, the slope is -1/2.

To find the slope of the line that is perpendicular to this, we need to take the negative reciprocal of -1/2. The negative reciprocal of -1/2 is 2/1 or simply 2.

Now, let's find the point of tangency on the curve g(x) = 3x^5 + 2x - 1. A line that is tangent to a curve at a specific point shares the same x-coordinate and y-coordinate as that point.

To find the point of tangency, we need to find the x-coordinate where the derivative of g(x) is equal to the slope we found earlier (2). So, let's take the derivative of g(x):

g'(x) = 15x^4 + 2

Now, let's set g'(x) equal to 2 and solve for x:

15x^4 + 2 = 2
15x^4 = 0
x^4 = 0

Taking the fourth root of both sides, we get:

x = 0

So, the point of tangency on the curve g(x) is (0, g(0)).

Now, we have the slope (2) and the point of tangency (0, g(0)). To find the equation of the line in slope-intercept form (y = mx + b), we can use the point-slope form:

y - y1 = m(x - x1)

Substituting the values we found, we have:

y - g(0) = 2(x - 0)
y - (3(0)^5 + 2(0) - 1) = 2x
y - 1 = 2x

Finally, rearranging the equation into slope-intercept form, we get:

y = 2x + 1

Therefore, the equation of the line that is perpendicular to y = -x/2 + 1 and tangent to g(x) = 3x^5 + 2x - 1 is y = 2x + 1 in slope-intercept form.