Comprehensive IGNOU BMTC-131 Solved Assignment 2024 for B.Sc (G) CBCS Students

IGNOU BMTC-131 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BMTC-131 Assignment Question Paper 2024

bmtc-131-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

bmtc-131-solved-assignment-2024-10cd7422-4b11-4531-9171-a6445826d855

BMTC-131 Solved Assignment 2024
  1. State whether the following statements are True or False? Justify your answers with the help of a short proof or a counter example:
a) The function f f fff, defined by f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x)=\cos x+\sin xf(x)=cos⁡x+sin⁡x, is an odd function.
b) d d x [ ∫ 2 e x ln ⁡ t d t ] = x − ln ⁡ 2 d d x ∫ 2 e x   ln ⁡ t d t = x − ln ⁡ 2 (d)/(dx)[int_(2)^(e^(x))ln tdt]=x-ln 2\frac{\mathrm{d}}{\mathrm{dx}}\left[\int_2^{\mathrm{e}^{\mathrm{x}}} \ln \mathrm{tdt}\right]=\mathrm{x}-\ln 2ddx[∫2exln⁡tdt]=x−ln⁡2.
c) The function f f fff, defined by f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x)=|x-2|f(x)=|x−2|, is differentiable in [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1].
d) y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y=x^2-3 x^3y=x2−3x3 has no points of inflection.
e) y = − x 2 y = − x 2 y=-x^(2)y=-x^2y=−x2 is increasing in [ − 5 , − 3 ] [ − 5 , − 3 ] [-5,-3][-5,-3][−5,−3].
  1. a) Find d y d x d y d x (dy)/(dx)\frac{d y}{d x}dydx, if y = x sin − 1 ⁡ x + 1 − x 2 y = x sin − 1 ⁡ x + 1 − x 2 y=xsin^(-1)x+sqrt(1-x^(2))y=x \sin ^{-1} x+\sqrt{1-x^2}y=xsin−1⁡x+1−x2.
b) Evaluate ∫ 0 π x sin ⁡ x 1 + cos 2 ⁡ x d x ∫ 0 π   x sin ⁡ x 1 + cos 2 ⁡ x d x int_(0)^(pi)(x sin x)/(1+cos^(2)x)dx\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x∫0πxsin⁡x1+cos2⁡xdx.
c) Find lim x → 1 x 3 − 3 x + 2 x 2 − 5 x + 4 lim x → 1   x 3 − 3 x + 2 x 2 − 5 x + 4 lim_(x rarr1)(x^(3)-3x+2)/(x^(2)-5x+4)\lim _{x \rightarrow 1} \frac{x^3-3 x+2}{x^2-5 x+4}limx→1x3−3x+2x2−5x+4.
  1. Trace the curve y 2 = x 2 ( x + 1 ) y 2 = x 2 ( x + 1 ) y^(2)=x^(2)(x+1)y^2=x^2(x+1)y2=x2(x+1) by the stating all the properties used to trace it.
  2. a) Find the length of the curve given by x = t 2 , y = 2 t 2 x = t 2 , y = 2 t 2 x=t^(2),y=2t^(2)x=t^2, y=2 t^2x=t2,y=2t2 in 0 ≤ t ≤ 2 0 ≤ t ≤ 2 0 <= t <= 20 \leq t \leq 20≤t≤2.
b) Find the angle between the curves y 2 = a x y 2 = a x y^(2)=axy^2=a xy2=ax and a y 2 = x 3 ( a > 0 ) a y 2 = x 3 ( a > 0 ) ay^(2)=x^(3)(a > 0)a y^2=x^3(a>0)ay2=x3(a>0), at the points of intersection other than the origin.
a) Evaluate ∫ x 2 d x ( x − 3 ) ( x − 5 ) ( x − 7 ) ∫ x 2 d x ( x − 3 ) ( x − 5 ) ( x − 7 ) int(x^(2)dx)/((x-3)(x-5)(x-7))\int \frac{x^2 d x}{(x-3)(x-5)(x-7)}∫x2dx(x−3)(x−5)(x−7)
b) Use Simpson’s method to approximate ∫ 0 8 ( x 2 − x + 3 ) d x ∫ 0 8   x 2 − x + 3 d x int_(0)^(8)(x^(2)-x+3)dx\int_0^8\left(x^2-x+3\right) d x∫08(x2−x+3)dx with 8 sub-intervals.
c) Find the derivatives of ln ⁡ ( 1 + x 2 ) ln ⁡ 1 + x 2 ln(1+x^(2))\ln \left(1+x^2\right)ln⁡(1+x2) w.r.t. tan − 1 ⁡ x tan − 1 ⁡ x tan^(-1)x\tan ^{-1} xtan−1⁡x.
  1. a) The curve a y 2 = x ( x − a ) 2 , a > 0 a y 2 = x ( x − a ) 2 , a > 0 ay^(2)=x(x-a)^(2),a > 0a y^2=x(x-a)^2, a>0ay2=x(x−a)2,a>0 has a loop between x = 0 x = 0 x=0x=0x=0 and x = a x = a x=ax=ax=a. Find the area of this loop.
b) Obtain the largest possible domain, and corresponding range, of the function f f fff, defined by f ( x ) = x − 2 3 − x f ( x ) = x − 2 3 − x f(x)=(x-2)/(3-x)f(x)=\frac{x-2}{3-x}f(x)=x−23−x.
c) Expand e 2 x e 2 x e^(2x)e^{2 x}e2x in powers of ( x − 1 ) ( x − 1 ) (x-1)(x-1)(x−1), up to four terms.
  1. a) Verify Rolle’s theorem for the function f f fff, defined by f ( x ) = x ( x − 2 ) e − x f ( x ) = x ( x − 2 ) e − x f(x)=x(x-2)e^(-x)f(x)=x(x-2) e^{-x}f(x)=x(x−2)e−x, on the interval [ 0 , 2 ] [ 0 , 2 ] [0,2][0,2][0,2].
b) Is the function f f fff, defined by
f ( x ) = x 2 − 5 x + 4 x 2 − 16 , x ≠ 4 f ( x ) = x 2 − 5 x + 4 x 2 − 16 , x ≠ 4 f(x)=(x^(2)-5x+4)/(x^(2)-16),x!=4f(x)=\frac{x^2-5 x+4}{x^2-16}, x \neq 4f(x)=x2−5x+4x2−16,x≠4
f ( 4 ) = 0 f ( 4 ) = 0 f(4)=0f(4)=0f(4)=0, continuous at x = 4 x = 4 x=4x=4x=4 ? Give reasons for your answer.
c) Evaluate ∫ 0 1 x 2 e 3 x d x ∫ 0 1   x 2 e 3 x d x int_(0)^(1)x^(2)e^(3x)dx\int_0^1 x^2 e^{3 x} d x∫01x2e3xdx
  1. a) Using Trapezoidal rule, calculate ∫ 0 1 d x 1 + x 2 ∫ 0 1   d x 1 + x 2 int_(0)^(1)(dx)/(1+x^(2))\int_0^1 \frac{d x}{1+x^2}∫01dx1+x2 by dividing the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] in 5 equal subintervals. Hence evaluate π π pi\piπ.
b) Find d y d x d y d x (dy)/(dx)\frac{d y}{d x}dydx, if y = x sin ⁡ x + ( sin ⁡ x ) x y = x sin ⁡ x + ( sin ⁡ x ) x y=x^(sin x)+(sin x)^(x)y=x^{\sin x}+(\sin x)^xy=xsin⁡x+(sin⁡x)x.
c) Find the perimeter of the cardioid r = a ( 1 − cos ⁡ θ ) r = a ( 1 − cos ⁡ θ ) r=a(1-cos theta)r=a(1-\cos \theta)r=a(1−cos⁡θ).
  1. a) If the first three non-zero terms of Maclaurin’s series for sin ⁡ x sin ⁡ x sin x\sin xsin⁡x are used to approximate sin ⁡ π / 2 sin ⁡ π / 2 sin pi//2\sin \pi / 2sin⁡π/2, show that the error is less than 1 / 50 1 / 50 1//501 / 501/50.
b) Find the least value of a 2 sec 2 ⁡ x + b 2 cosec 2 ⁡ x a 2 sec 2 ⁡ x + b 2 cosec 2 ⁡ x a^(2)sec^(2)x+b^(2)cosec^(2)xa^2 \sec ^2 x+b^2 \operatorname{cosec}^2 xa2sec2⁡x+b2cosec2⁡x, where a > 0 , b > 0 a > 0 , b > 0 a > 0,b > 0a>0, b>0a>0,b>0.
c) Evaluate lim x → 0 ( 1 + x ) 1 / x lim x → 0   ( 1 + x ) 1 / x lim_(x rarr0)(1+x)^(1//x)\lim _{x \rightarrow 0}(1+x)^{1 / x}limx→0(1+x)1/x
  1. a) Find the slope of the normal to the curve y = x 3 y = x 3 y=x^(3)y=x^3y=x3 at ( 1 2 , 1 8 ) 1 2 , 1 8 ((1)/(2),(1)/(8))\left(\frac{1}{2}, \frac{1}{8}\right)(12,18).
b) Find the points of inflexion of the curve y = a 2 x x 2 + a 2 y = a 2 x x 2 + a 2 y=(a^(2)x)/(x^(2)+a^(2))y=\frac{a^2 x}{x^2+a^2}y=a2xx2+a2. Also, show that they lie on a straight line.
c) Evaluate ∫ e x ⋅ x 2 − x + 1 ( 1 + x 2 ) 3 / 2 d x ∫ e x ⋅ x 2 − x + 1 1 + x 2 3 / 2 d x inte^(x)*(x^(2)-x+1)/((1+x^(2))^(3//2))dx\int e^x \cdot \frac{x^2-x+1}{\left(1+x^2\right)^{3 / 2}} d x∫ex⋅x2−x+1(1+x2)3/2dx
\(c=a\:cos\:B+b\:cos\:A\)

BMTC-131 Sample Solution 2024

bmtc-131-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

bmtc-131-solved-assignment-2024-ss-10cd7422-4b11-4531-9171-a6445826d855

BMTC-131 Solved Assignment 2024
  1. State whether the following statements are True or False? Justify your answers with the help of a short proof or a counter example:
a) The function f f fff, defined by f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x)=\cos x+\sin xf(x)=cos⁡x+sin⁡x, is an odd function.
Answer:
To determine whether the statement "The function f f fff, defined by f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x) = \cos x + \sin xf(x)=cos⁡x+sin⁡x, is an odd function" is true, we need to understand the definition of an odd function and then apply it to f ( x ) f ( x ) f(x)f(x)f(x).

Definition of an Odd Function

A function f ( x ) f ( x ) f(x)f(x)f(x) is said to be odd if it satisfies the condition f ( − x ) = − f ( x ) f ( − x ) = − f ( x ) f(-x)=-f(x)f(-x) = -f(x)f(−x)=−f(x) for all x x xxx in its domain. This means that the function is symmetric with respect to the origin.

Applying the Definition to f ( x ) f ( x ) f(x)f(x)f(x)

The given function is f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x) = \cos x + \sin xf(x)=cos⁡x+sin⁡x. To check if it’s odd, we compute f ( − x ) f ( − x ) f(-x)f(-x)f(−x) and compare it with − f ( x ) − f ( x ) -f(x)-f(x)−f(x).
  1. Compute f ( − x ) f ( − x ) f(-x)f(-x)f(−x):
    f ( − x ) = cos ⁡ ( − x ) + sin ⁡ ( − x ) f ( − x ) = cos ⁡ ( − x ) + sin ⁡ ( − x ) f(-x)=cos(-x)+sin(-x)f(-x) = \cos(-x) + \sin(-x)f(−x)=cos⁡(−x)+sin⁡(−x)
    Using the even property of cosine and the odd property of sine:
    f ( − x ) = cos ⁡ x − sin ⁡ x f ( − x ) = cos ⁡ x − sin ⁡ x f(-x)=cos x-sin xf(-x) = \cos x – \sin xf(−x)=cos⁡x−sin⁡x
  2. Compare f ( − x ) f ( − x ) f(-x)f(-x)f(−x) with − f ( x ) − f ( x ) -f(x)-f(x)−f(x):
    − f ( x ) = − ( cos ⁡ x + sin ⁡ x ) = − cos ⁡ x − sin ⁡ x − f ( x ) = − ( cos ⁡ x + sin ⁡ x ) = − cos ⁡ x − sin ⁡ x -f(x)=-(cos x+sin x)=-cos x-sin x-f(x) = -(\cos x + \sin x) = -\cos x – \sin x−f(x)=−(cos⁡x+sin⁡x)=−cos⁡x−sin⁡x
    Clearly, f ( − x ) = cos ⁡ x − sin ⁡ x f ( − x ) = cos ⁡ x − sin ⁡ x f(-x)=cos x-sin xf(-x) = \cos x – \sin xf(−x)=cos⁡x−sin⁡x is not equal to − f ( x ) = − cos ⁡ x − sin ⁡ x − f ( x ) = − cos ⁡ x − sin ⁡ x -f(x)=-cos x-sin x-f(x) = -\cos x – \sin x−f(x)=−cos⁡x−sin⁡x.

Conclusion

Since f ( − x ) ≠ − f ( x ) f ( − x ) ≠ − f ( x ) f(-x)!=-f(x)f(-x) \neq -f(x)f(−x)≠−f(x) for the function f ( x ) = cos ⁡ x + sin ⁡ x f ( x ) = cos ⁡ x + sin ⁡ x f(x)=cos x+sin xf(x) = \cos x + \sin xf(x)=cos⁡x+sin⁡x, the function is not an odd function. Therefore, the statement is false.
b) d d x [ ∫ 2 e x ln ⁡ t d t ] = x − ln ⁡ 2 d d x ∫ 2 e x   ln ⁡ t d t = x − ln ⁡ 2 (d)/(dx)[int_(2)^(e^(x))ln tdt]=x-ln 2\frac{\mathrm{d}}{\mathrm{dx}}\left[\int_2^{\mathrm{e}^{\mathrm{x}}} \ln \mathrm{tdt}\right]=\mathrm{x}-\ln 2ddx[∫2exln⁡tdt]=x−ln⁡2.
Answer:
To determine the truth of the statement
d d x [ ∫ 2 e x ln ⁡ t d t ] = x − ln ⁡ 2 , d d x ∫ 2 e x   ln ⁡ t d t = x − ln ⁡ 2 , (d)/(dx)[int_(2)^(e^(x))ln tdt]=x-ln 2,\frac{d}{dx}\left[\int_2^{e^x} \ln t \, dt\right] = x – \ln 2,ddx[∫2exln⁡tdt]=x−ln⁡2,
we need to use the Fundamental Theorem of Calculus and the chain rule.

Fundamental Theorem of Calculus and Chain Rule

The Fundamental Theorem of Calculus states that if F ( x ) F ( x ) F(x)F(x)F(x) is an antiderivative of f ( x ) f ( x ) f(x)f(x)f(x), then
d d x [ ∫ a g ( x ) f ( t ) d t ] = f ( g ( x ) ) ⋅ g ′ ( x ) . d d x ∫ a g ( x )   f ( t ) d t = f ( g ( x ) ) ⋅ g ′ ( x ) . (d)/(dx)[int_(a)^(g(x))f(t)dt]=f(g(x))*g^(‘)(x).\frac{d}{dx}\left[\int_a^{g(x)} f(t) \, dt\right] = f(g(x)) \cdot g'(x).ddx[∫ag(x)f(t)dt]=f(g(x))⋅g′(x).
In this case, f ( t ) = ln ⁡ t f ( t ) = ln ⁡ t f(t)=ln tf(t) = \ln tf(t)=ln⁡t and g ( x ) = e x g ( x ) = e x g(x)=e^(x)g(x) = e^xg(x)=ex. The derivative of g ( x ) = e x g ( x ) = e x g(x)=e^(x)g(x) = e^xg(x)=ex is g ′ ( x ) = e x g ′ ( x ) = e x g^(‘)(x)=e^(x)g'(x) = e^xg′(x)=ex.

Applying the Theorem

Applying the theorem to the given integral, we get:
d d x [ ∫ 2 e x ln ⁡ t d t ] = ln ⁡ ( e x ) ⋅ e x . d d x ∫ 2 e x   ln ⁡ t d t = ln ⁡ ( e x ) ⋅ e x . (d)/(dx)[int_(2)^(e^(x))ln tdt]=ln(e^(x))*e^(x).\frac{d}{dx}\left[\int_2^{e^x} \ln t \, dt\right] = \ln(e^x) \cdot e^x.ddx[∫2exln⁡tdt]=ln⁡(ex)⋅ex.
Since ln ⁡ ( e x ) = x ln ⁡ ( e x ) = x ln(e^(x))=x\ln(e^x) = xln⁡(ex)=x, this simplifies to:
x ⋅ e x . x ⋅ e x . x*e^(x).x \cdot e^x.x⋅ex.

Comparing with the Given Statement

The given statement is:
d d x [ ∫ 2 e x ln ⁡ t d t ] = x − ln ⁡ 2. d d x ∫ 2 e x   ln ⁡ t d t = x − ln ⁡ 2. (d)/(dx)[int_(2)^(e^(x))ln tdt]=x-ln 2.\frac{d}{dx}\left[\int_2^{e^x} \ln t \, dt\right] = x – \ln 2.ddx[∫2exln⁡tdt]=x−ln⁡2.
However, from our calculation, we found that the derivative is actually x ⋅ e x x ⋅ e x x*e^(x)x \cdot e^xx⋅ex, not x − ln ⁡ 2 x − ln ⁡ 2 x-ln 2x – \ln 2x−ln⁡2.

Conclusion

The statement is false. The correct derivative of the integral ∫ 2 e x ln ⁡ t d t ∫ 2 e x   ln ⁡ t d t int_(2)^(e^(x))ln tdt\int_2^{e^x} \ln t \, dt∫2exln⁡tdt with respect to x x xxx is x ⋅ e x x ⋅ e x x*e^(x)x \cdot e^xx⋅ex, not x − ln ⁡ 2 x − ln ⁡ 2 x-ln 2x – \ln 2x−ln⁡2.
c) The function f f fff, defined by f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x)=|x-2|f(x)=|x−2|, is differentiable in [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1].
Answer:
To determine whether the statement "The function f f fff, defined by f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2|, is differentiable in [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1]" is true, we need to examine the differentiability of the function f ( x ) f ( x ) f(x)f(x)f(x) within the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1].

Definition of Differentiability

A function is differentiable at a point if its derivative exists at that point. For a function to be differentiable on an interval, it must be differentiable at every point in that interval.

Analyzing the Function f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2|

The function f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2| can be expressed as:
  • f ( x ) = 2 − x f ( x ) = 2 − x f(x)=2-xf(x) = 2 – xf(x)=2−x for x ≤ 2 x ≤ 2 x <= 2x \leq 2x≤2
  • f ( x ) = x − 2 f ( x ) = x − 2 f(x)=x-2f(x) = x – 2f(x)=x−2 for x > 2 x > 2 x > 2x > 2x>2
In the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1], f ( x ) = 2 − x f ( x ) = 2 − x f(x)=2-xf(x) = 2 – xf(x)=2−x, since all values of x x xxx in this interval are less than 2.

Differentiability in [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1]

For x ∈ [ 0 , 1 ] x ∈ [ 0 , 1 ] x in[0,1]x \in [0, 1]x∈[0,1], f ( x ) = 2 − x f ( x ) = 2 − x f(x)=2-xf(x) = 2 – xf(x)=2−x, which is a linear function. The derivative of f ( x ) f ( x ) f(x)f(x)f(x) in this interval is f ′ ( x ) = − 1 f ′ ( x ) = − 1 f^(‘)(x)=-1f'(x) = -1f′(x)=−1, which exists for all x x xxx in [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1].

Conclusion

Since the derivative of f ( x ) = | x − 2 | f ( x ) = | x − 2 | f(x)=|x-2|f(x) = |x – 2|f(x)=|x−2| exists and is constant ( − 1 − 1 -1-1−1) throughout the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1], the function f f fff is differentiable in [ 0 , 1 ] [ 0 , 1 ] [0,1][0, 1][0,1]. Therefore, the statement is true.
d) y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y=x^2-3 x^3y=x2−3x3 has no points of inflection.
Answer:
To determine whether the function y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y = x^2 – 3x^3y=x2−3x3 has any points of inflection, we need to check the concavity of the function. A point of inflection occurs when the concavity of the function changes from concave up to concave down or vice versa. This change in concavity is characterized by the second derivative of the function.
A point of inflection occurs at x = c x = c x=cx = cx=c if and only if the second derivative y ″ ( c ) y ″ ( c ) y^(″)(c)y”(c)y″(c) changes sign at that point. Specifically, y ″ ( c ) = 0 y ″ ( c ) = 0 y^(″)(c)=0y”(c) = 0y″(c)=0 and the sign of y ″ ( x ) y ″ ( x ) y^(″)(x)y”(x)y″(x) changes as x x xxx crosses c c ccc.
Let’s find the first and second derivatives of the function y = x 2 − 3 x 3 y = x 2 − 3 x 3 y=x^(2)-3x^(3)y = x^2 – 3x^3y=x2−3x3:
  1. First derivative:
    y ′ = 2 x − 9 x 2 y ′ = 2 x − 9 x 2 y^(‘)=2x-9x^(2)y’ = 2x – 9x^2y′=2x−9x2
  2. Second derivative:
    y ″ = 2 − 18 x y ″ = 2 − 18 x y^(″)=2-18 xy” = 2 – 18xy″=2−18x
Now, we need to find the values of x x xxx where y ″ ( x ) = 0 y ″ ( x ) = 0 y^(″)(x)=0y”(x) = 0y″(x)=0:
2 − 18 x = 0 2 − 18 x = 0 2-18 x=02 – 18x = 02−18x=0
Solving for x x xxx:
18 x = 2 ⟹ x = 2 18 = 1 9 18 x = 2 ⟹ x = 2 18 = 1 9 18 x=2Longrightarrowx=(2)/(18)=(1)/(9)18x = 2 \implies x = \frac{2}{18} = \frac{1}{9}18x=2⟹x=218=19
So, y ″ ( x ) = 0 y ″ ( x ) = 0 y^(″)(x)=0y”(x) = 0y″(x)=0 at x = 1 9 x = 1 9 x=(1)/(9)x = \frac{1}{9}x=19.
Now, we need to determine the sign of y ″ ( x ) y ″ ( x ) y^(″)(x)y”(x)y″(x) in the intervals around x = 1 9 x = 1 9 x=(1)/(9)x = \frac{1}{9}x=19:
  • For x < 1 9 x < 1 9 x < (1)/(9)x < \frac{1}{9}x<19, y ″ ( x ) = 2 − 18 x > 0 y ″ ( x ) = 2 − 18 x > 0 y^(″)(x)=2-18 x > 0y”(x) = 2 – 18x > 0y″(x)=2−18x>0 (since x x xxx is positive).
  • For x > 1 9 x > 1 9 x > (1)/(9)x > \frac{1}{9}