..READ THIS FIRST..Problems
about maxima, minima, inflection points, ...
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Problems about maxima, minima, inflection points, ...
Level 2 problems
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Prove that there is a real number r such that _______ | a + x 1 x arctan( | ----- ) = - . arcsin(-) + r \| a - x 2 a The number a is a real strictly positive number.To prove this, we'll show that both sides have the same derivate. _______ d | a + x 1 1/2 2a -- arctan( | ----- ) = -----------.-----------.----------- dx \| a - x a + x _______ 2 1 + ----- | a + x (a - x) a - x | ----- \| a - x 1/2 = ... = ----------- _________ V a2 - x2 and d 1 x 1/2 . 1/a -- ( - . arcsin(-) ) = ---------------- dx 2 a ____________ V 1- (x/a)2 1/2 = ------------ __________ V a2 - x2 -
Calculate the condition for the real value of the parameter h such that the following function has a minimum and a maximum. h + 3x - x2 f(x) = --------------- x - 4f(x) is not defined for x = 4. Now we assume x is not 4. The derivative of the function is - x2 + 8x - 12 - h f'(x) = ------------------- (x - 4)(x - 4) f'(x) = 0 <=> - x2 + 8x - 12 - h = 0 The discriminant D is 4(4-h) If D < 0 , or h > 4, f'(x) is negative for all x and there is no maximum and no minimum. If D > 0 , or h < 4, f'(x) has two roots and goes from negative to positive and to negative again. In that case, the function has a maximum and a minimum. -
Recall previous problem and take h < 4. Investigate the number of intersection points of f(x) with the line y = k. Deduce from this the difference between the maximum and minimum value of f(x).
The x values of the intersection points are the solutions ofk(x - 4) = h + 3x - x2 <=> x2 + (k - 3)x - 4k - h From this, we see that there are at most two intersection points. The number of intersection points depends on the sign of D = (k - 3)2 + 16k + 4h = k2 + 10k + 9 + 4h D = 0 for k = -5 + 2.sqrt(4 - h) and for k = -5 - 2.sqrt(4 - h) For these k values there is just one intersection point. For k in ]-5 - 2.sqrt(4 - h); -5 + 2.sqrt(4 - h)[ there are no intersection points. For k < -5 - 2.sqrt(4 - h) or k > -5 + 2.sqrt(4 - h) there are twointersection points. From this we deduce that the maximum value of f(x) =-5 + 2.sqrt(4 - h). The minimum value is -5 - 2.sqrt(4 - h). The difference between maximum and minimum is 4.sqrt(4-h). -
Calculate the inflection points of y = (x + 1)/(x2 + 1) -x2 - 2x + 1 y' = ... = -------------- (x2 + 1 )2 2x3 + 6x2 - 6x - 2 y" = ... = --------------------- (x2 + 1 )3 2(x - 1)(x+2-sqrt(3))(x+2+sqrt(3)) y" = ... = ---------------------------------- (x2 + 1 )3 The inflection points are (1;1) (-2+sqrt(3);(1+sqrt(3))/4) (-2-sqrt(3);(1-sqrt(3))/4) -
Calculate the inflection points of y = sin(2x)+3sin(2x/3) in [0,3pi/2]. 3 hint : sin(3t) = 3.sin(t) - 4 sin (t)y' = 2cos(2x)+2cos(2x/3) y" = -4sin(2x) - (4/3)sin(2x/3) y" = 0 <=> sin(2x) + (1/3)sin(2x/3) =0 <=> 3.sin(2x/3) - 4sin3(2x/3) + (1/3)sin(2x/3) = 0 <=> sin(2x/3) . (10 - 12.sin2(2x/3)) = 0 <=> sin(2x/3) = 0 or sin2(2x/3) = 5/6 The inflection points are (0;3pi/2) (1.725 ;2.43) and (2.986;2.43) -
Calculate the inflection points of y = e2x - 5 ex +6
y' = 2.e2x - 5 ex y" = 4.e2x - 5 ex y" = 0 <=> ex= 5/4 <=> x = 0.22 Inflection point (0.223 ; 1.31) -
Study the curve f(x) = (x2 - x)2 Give the number of intersection points of that curve with the line y = m.
f'(x) = 2x(x - 1)(2x - 1) f"(x) = 2(6x.x -6x + 1) x | 0 1/2 1 ------------------------------------------ f'(x) | - 0 + 0 - 0 + ------------------------------------------ f(x) | 0 1/16 0There is a minimum in (0,0) and (1,0) and a maximum in (3/2,1/16).
f"(x) = 0 <=> x = 1/2 + 1/sqrt(12) or x = 1/2 - 1/sqrt(12)
For these x-values, we have the inflection points.
From the table, we see that the curve is never beneath the x-axis.
So, the line y = m has no intersection points for m < 0.
For m = 0, there are 2 intersection points.
For 0 < m < 1/16 there are are 4 intersection points.
For m = 1/16 there are 3 intersection points.
For m > 1/16 there are 2 intersection points.
The number of intersection points is the number of roots of the equation(x2 - x)2 - m = 0
Level 3 problems
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The curve F is a circle with radius r and center O. AB is a central line with points A and B on F. The line CD is parallel to AB with points C and D on F, such that ACDB is a isosceles trapezium ( |AC| = |DB| ). The angle(DOC) = 2t radians.
Calculate the area of the trapezium as a function of t.
Calculate the value of t such that this area is maximum.
Take x-axis and y-axis orthogonal with origin O, such that A and B are on y. Then point C has coordinates (r.cos(t), r.sin(t)) .
|AB| = 2r ; |CD| = 2r.sin(t) ; height = r.cos(t)
The area of the trapezium is (2r + 2r.sin(t)).r.cos(t)/2
= r2.(cos(t) + sin(t).cos(t))The derivative = r2.( -sin(t) + cos2(t) - sin2(t) ) = r2.( -sin(t) + 1- 2.sin2(t) ) Since t is in [0,pi/2], sin(t) is positive and the only value that makes the derivative change sign is sin(t) = 1/2 <=> t = pi/6. The trapezium is maximum for t=pi/6. Then the trapezium is half a regular hexagon. -
A function y = f(x) is implicitly defined by y -x - sqrt(y.y + 2(x - 1)y + 4x) = 0 - Calculate f(x)
- What is the domain of the function f
- Determine the maxima and minima
y -x = sqrt(y.y + 2(x - 1)y + 4x) <=> (y -x)2 = y2 + 2(x - 1)y + 4x with (y-x) >= 0 <=> ... x2 - 4x <=> y = --------- with y >= x 4x + 2 x is in domain of f <=> x is not -1/2 and y >= x x2 - 4x <=> x is not -1/2 and --------- - x >= 0 4x + 2 <=> ... x(x + 2) <=> x is not -1/2 and --------- =< 0 (2x + 1) Investigation of sign gives x | -2 -1/2 0 ------------------------------- | - 0 + | - 0 + So the domain f is ]-infty ; -2] U ]-1/2 ; 0 ] (x + 2)(x - 1) y' = ... = --------------- (2x + 1)(2x + 1) Investigation of sign gives x | -2 -1/2 1 ------------------------------- y' | + 0 - | - 0 + So the function increases in ]-infty ; -2] and decreases in ]-1/2 ; 0 ]. The image is relative maximum for x = -2. Then y = -2. The image is relative minimum for x = 0. Then y = 0. -
Define all real values of m such that the asymptotes of the curve 2(m-1)x - m + 1 y = ---------------- (m+3)x + m intersect in a point above the line y = 2x-1-m The vertical asymptote is x = ------- m+3 2(m-1) The horizontal asymptote is y = ---------- m+3 Now, we must have y > 2x - 1 2(m-1) -2m <=> ------ > ---- -1 m+3 m+3 <=> ... 5m + 1 <=> --------- > 0 m + 3 <=> (5m + 1)(m + 3) > 0 The m values are defined by ( m <-3 or m > 1/5 ) -
Investigate if there is a vertical asymptote at x = 0 for the function arcsin(2x) - 2 arcsin(x) y = ------------------------- x3We have to calculate the limit of f(x) for x -> 0 arcsin(2x) - 2 arcsin(x) lim ------------------------- 0 x3 This gives the case 0/0 . We use L'Hospitals rule. 2 / sqrt(1 - 4 x2) - 1 / sqrt(1 - x2) = lim ----------------------------------------- 0 3 x2 _______ _________ | 2 | 2 2 \| 1 - x - \| 1 - 4 x = lim -------------------------. ----------------------------- 0 _________ _______ | 2 | 2 2 3\| 1 - 4 x \| 1 - x x The first factor has limit 2/3 . On the second factor we use L'Hospitals rule. -x /sqrt(1 - x2) + 4 x /sqrt(1 - 4 x2) = (2/3).lim ------------------------------------------- 0 2 x -1 /sqrt(1 - x2) + 4 /sqrt(1 - 4 x2) = (2/3).lim ------------------------------------------- 0 2 = 1 So, there is no vertical asymptote at x = 0. -
Calculate the horizontal asymptote to the function sin(1/x) y = ---------------- arctan(1/x)We have to calculate sin(1/x) lim ---------------- infty arctan(1/x) This gives the case 0/0. We can use L'Hospitals rule, but first we can simplify the limit by the substitution 1/x = t Then we have sin(t) = lim ------------ 0 arctan(t) Now, we use L'Hospitals rule. cos(t) = lim -------------------- = 1 0 1 / sqrt(1 + t2) So, y = 1 is a horizontal asymptote
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