Question #108748
Which of the following statements are true? Give reasons for your answers, in the
form of a short proof or a counterexample.
i)
2
2
2
dx
dy
dx
d y






=
ii) The inverse function of 3x
y = e is ln x
3
1
y = .
iii) If f is increasing and 0 )x(f > on an interval I, then
)x(f
1
)x(g = is decreasing on I.
iv) An equation of the tangent line to the parabola 2
y = x at )4,2 (− is
y − 4 = 2 x(x + )2 .
v) If f is one-one onto and differentiable on R , then
f )6(
1
f( )6()'
1

=

.
1
Expert's answer
2020-04-09T14:55:04-0400

1.


d2ydx2=(dydx)2,False{d^2y\over dx^2}=({dy \over dx})^2, False

Let y=x.y=x. Then


dydx=1,d2ydx2=0{dy \over dx}=1, {d^2y\over dx^2}=0d2ydx2=(dydx)2{d^2y\over dx^2}=({dy \over dx})^2d2ydx2=01=(dydx)2{d^2y\over dx^2}=0\not=1=({dy \over dx})^2

2. The inverse function of y=e3xy=e^{3x} is y=13lnx.y=\dfrac{1}{3}\ln x. False

If x=1:e3(1)=e3.x=-1: e^{3(-1)}=e^{-3}.

But the function y=13lnxy=\dfrac{1}{3}\ln x is undefined at x=1.x=-1.

3.  If ff is increasing and f(x)>0f(x)>0 on an interval I,I, then g(x)=1/f(x)g(x)=1/f(x) is deceasing on I.I.

True

f(x)>0f(x)>0 on I, f(x)f(x) is incresing on II

Let x1I,x2Ix_1 \in I, x_2\in I

For x2>x1,f(x2)>f(x1)>0For \ x_2>x_1, f(x_2)>f(x_1)>0

Hence

For x2>x1,0<1/f(x2)<1/f(x1)For \ x_2>x_1, 0<1/f(x_2)<1/f(x_1)

This means that g(x)=1/f(x)g(x)=1/f(x) is deceasing on I.I.

4. An equation of the tangent line to the parabola y=x2y=x^2 at (2,4)(-2,4) is y=2x(x+2)y=2x(x+2)

False. The equation of the line is y=kx+by=kx+b


k=y(2)=2(2)=4k=y'(-2)=2(-2)=-4y=4x+by=-4x+by(2)=4(2)+b=4=>b=4y(-2)=-4(-2)+b=4=>b=-4

The equation of the tangent line to th parabola y=x2y=x^2 at (2,4)(-2, 4) is


y=4x4y=-4x-4

5.  If ff is one - one onto and differentiable on R, then


(f1)(6)=1/f(6)(f^{-1})'(6)=1/f'(6)

False

Let y=x3y=x^3

y=3x2,y(6)=3(6)2=108y'=3x^2, y'(6)=3(6)^2=108


y1=x3y^{-1}=\sqrt[3]{x}

(y1)=2x23(y^{-1})'=\dfrac{2}{\sqrt[3]{x^2}}

(y1)(6)=2623=6331108=1y(6)(y^{-1})'(6)=\dfrac{2}{\sqrt[3]{6^2}}=\dfrac{\sqrt[3]{6}}{3}\not=\dfrac{1}{108}=\dfrac{1}{y'(6)}



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