Answer to Question #108748 in Calculus for Preeti

Question

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
′
=
−
.

Answer 1

1.

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

Let $y=x.$ Then

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

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

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

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

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

True

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

Let $x_1 \in I, x_2\in I$

$For \ x_2>x_1, f(x_2)>f(x_1)>0$

Hence

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

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

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

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

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

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

$$y=-4x-4$$

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

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

False

Let $y=x^3$

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

$y^{-1}=\sqrt[3]{x}$

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

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