Q1 Can you pass this 3 question AP calculus Quiz in 10 minutes

[karush]

1. $f(x)=(2x+1)^3$ and let g be the inverse function of f. Given that$f(0)=1$ what is the value of $g'(1)$?

A $-\dfrac{2}{27}$ B $\dfrac{1}{54}$ C $\dfrac{1}{27}$ D $\dfrac{1}{6}$ E 6

Spoiler: solution

$f(x) = (2x+1)^3 \implies f'(x) = 6(2x+1)^2$
$f$ and $g$ are inverses tells us two things ...
(1) $f(0) = 1 \implies g(1)=0$
(2) $f[g(x)] = x$
take the derivative of equation (2) ...
$f'[g(x)] \cdot g'(x) = 1 \implies g'(x) = \dfrac{1}{f'[g(x)]}$
$g'(1)=\dfrac{1}{f'[g(1)]}
= \dfrac{1}{f'(0)}
= \dfrac{1}{6}$

2. given that $\left[f(x)=x-2,\quad g(x)=\dfrac{x}{x^2+1}\right]$
find $f(g(-2))$
A $\dfrac{-11}{5}\quad $ B $\dfrac{-4}{17}\quad$ C $-3\quad $ D $\dfrac{14}{85}\quad$ E $\dfrac{-12}{5}$

Spoiler: 2. solution

find $g(-2)$
$$\dfrac{-2}{(2)^2+1}=\dfrac{-2}{5}$$
then solve $f(-2/5)$
$$\dfrac{-2}{5}-2
=\dfrac{-2}{5}-\dfrac{-10}{5}
=\dfrac{-12}{5}$$

3. The function f is defined by $f(x)=\dfrac{x}{x+2}$ What points $(x,y)$ on the graph of $f$ have the property that the line tangent to $f$ at $(x,y)$ has slope $\dfrac{1}{2}$?
$$\textsf{A (0,0) only B $\left(\dfrac{1}{2},\dfrac{1}{5}\right)$ only
C (0,0) and (-4,2)
D (0,0)and $\left(4,\dfrac{2}{3}\right)$
E DNE}$$

Spoiler: 3. solution

$\dfrac{d}{dx}\dfrac{x}{x+2}=\dfrac{1}{2}
\quad x=-4,0 \quad\therefore y=2,0

Note solutions came by replies from MHB forum

 

[HOI]

1) g(x) is the inverse function to $f(x)= (2x+1)^3$. The problem asks you to find g'(1). First note that $f(0)= (2(0)+ 1)^3= 1^3= 1$ so g(1)= 0. Further, we learn in Calculus that the derivative of the inverse of f(x) is 1 over the derivative of f. Since $f(x)= (2x+ 1)^3$, $f'(x)= 3(2x+1)^2(2)= 6(2x+1)^2$. $f'(0)= 6(1)= 6$ so $g'(1)= \frac{1}{6}$.

The HARD way to do this, for those who like to do things the hard way, is to actually find g! If $y= (2x+ 1)^3$ then $2x+1= y^{1/3}$, $2x= y^{1/3}-1$, $x= \frac{y^{1/3}- 1}{2}$, so $g(x)= \frac{x^{1/3}- 1}{2}$. So $g'(x)= \frac{1}{6}x^{-2/3}$ and $g'(1)= \frac{1}{6}$. Wow, I got the same answer!

 

[HOI]

2. given that f(x)=x−2, $g(x)= \frac{x}{x^2+ 1}$find
f(g(−2)).
Well, that's easy- more arithmetic than Calculus! $g(-2)= \frac{-2}{(-2)^2+ 1}= \frac{-2}{4+ 1}= -\frac{2}{5}$. Then f(g(-2))= f(-2/5)= -2/5- 2= -2/5- 10/5= -12/5.

 

[HOI]

3. The function f is defined by $f(x)= \frac{x}{x+2}$.
What points (x,y) on the graph of fhave the property that the line tangent to f at (x,y) has slope $\frac{1}{2}$?

In Calculus you learn that the slope of the tangent line is the same as the derivative. The derivative is $f'= \frac{(x+ 2)- x}{(x+2)^2}= \frac{2}{(x+2)^2}$. So the question is "What x satisfies $\frac{2}{(x+2)^2}= \frac{1}{2}$. Multiply both sides by $2(x+2)^2$ to get $\frac{4}= (x+2)^2$. Taking the square root or both sides, $\pm 2= x+ 2$. With the positive sign, that is 2= x+ 2 or x= 0. With the negative sign, that is -2= x+ 2 or x= -4. When x= 0, $y= \frac{0}{0+ 2}= 0$. When x= -4, $y= \frac{-4}{-4+ 2}= \frac{-4}{-2}= 2$. The points, (x, y), such that the slope of the tangent line is 1/2 are (0, 0) and (-4, 2).