Let f be a function such that f "(x) = 6x + 8.
find f(x) if the graph of f is tangent to the line
3x - y = 2 at the point (0,-2).

answer:   x = 3 + √11     and     x = 3 - √11

step-by-step explanation:

x² - 6x + 9 = 11

(x - 3)² = 11

x - 3 = ±√11

    x = 3 ± √11

x = 3 + √11     and     x = 3 - √11

region b will make the inequalities y < 2x + 1 and y > -x - 2 true ⇒ answer a

step-by-step explanation:

* lets study the graph and the inequality to solve the question

- the inequality y > -x - 2

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = -2 that means the line intersect the y-axis at point (0 , -2)

∵ the sign of the inequality is >

∴ we shad the part over the line

  region a and b

- the inequality y < 2x + 1

∵ the form of the inequality is y > mx + c , m is the slope of the line

  and c is the y-intercept (the line intersect y-axis at point (0 , c)

∴ c = 1 that means the line intersect the y-axis at point (0 , 1)

∵ the sign of the inequality is <

∴ we shad the part under the line

  region b and c

∵ the common region between the two lines is region b

∴ region b will make the inequalities y < 2x + 1 and y > -x - 2 true