Question

Which of the following statements are true? Give reasons for your answers, in the form of a short proof or a counter example. (1) d^2/dx^2=(dy/dx) ^2 (2) The inverse function of y=e^3x is y=1/3lnx. (3) If f is increasing and f(x) >0 on an interval I, then 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-4=2x(x+2). (5) If f is one - one onto and differentiable on R, then (f^-1) '(6) =1/f'(6)

Answer 1

1)It is false. For $y=x$ we have $\frac{d^2y}{dx^2}=0$ and $\left(\frac{dy}{dx}\right)^2=1$

2)It is true. If $y=e^{3x}$ , then $3x=\ln y$ and $x=\frac{1}{3}\ln y$. So inverse of $y=e^{3x}$ is $y=\frac{1}{3}\ln x$.

3)It is true. Take $x_1,x_2\in I$, where $x_1<x_2$. Since $f$ is increasing on $I$ and $f(x)>0$ on $I$, we have $0<f(x_1)<f(x_2)$, so $0<\frac{1}{f(x_2)}<\frac{1}{f(x_1)}$, that is $g(x_2)<g(x_1)$.

4)It is false, because $y-4=2x(x+2)$ is not an equation of straight line, so it is not a tangen line to curve.

5)It is not true. For $f(x)=x^3$ we have $f^{-1}(x)=\sqrt[3]{x}$. So $f'(x)=3x^2$ and $(f^{-1})'(x)=\frac{1}{3\sqrt[3]{x^2}}$.

$f'(6)=3\cdot 6^2=108$ and $(f^{-1})'(6)=\frac{1}{3\sqrt[3]{6^2}}=\frac{1}{3\sqrt[3]{36}}$, so $(f^{-1})'(6)\neq\frac{1}{f'(6)}$