# Math Help - AP style Calculus Questions

1. ## AP style Calculus Questions

1. Consider the differential equation dy/dx = 3(x^2) / e^2y

(a) Find a solution for y = f(x) to the differential equation satisfying f(0) = ½.

(b) Find the domain and range of the function f found in part (a). Justify your answer.

2. Suppose that the function f has a continuous second derivative for all x, and that f(0) = 2, f ‘ (0) = -3, and f “ (0) = 0. Let g be a function whose derivative is given by g'(x) = e^-2x(3f(x) + 2f'(x)) for all x.

(a) Write an equation of the line tangent to the graph of f at the point where x = 0.

(c) Given that g(0) = 4, write an equation of the line tangent to the graph of g at the point where x = 0.

(d) Show that g''(x) = e^-2x(-6f(x) 2f''(x)) At x = 0, is g ‘ (x) = 0 and g “ (x) < 0? Justify your answer.

I apologize for not thanking whoever it was that helped me with the last questions i had. I ended up checking my answers with my friend, and kind of forgot to check here.

2. Originally Posted by Blue Griffin
2. Suppose that the function f has a continuous second derivative for all x, and that f(0) = 2, f ‘ (0) = -3, and f “ (0) = 0. Let g be a function whose derivative is given by g'(x) = e^-2x(3f(x) + 2f'(x)) for all x.

(a) Write an equation of the line tangent to the graph of f at the point where x = 0.

(c) Given that g(0) = 4, write an equation of the line tangent to the graph of g at the point where x = 0.

(d) Show that g''(x) = e^-2x(-6f(x) 2f''(x)) At x = 0, is g ‘ (x) = 0 and g “ (x) < 0? Justify your answer.
For A with the info given $f(0)=2 \iff (0,2)$ and $m=f'(0)=-3$ so we get $y-2=-3(x-0) \iff y=-3x+2$

for c) using the equation

$g'(x)=e^{-2x}(3f(x)+2f'(x)) \mbox{ and } g(0)=4 \iff (0,4)$

so $g'(0)=e^{0}(3f(0)+2f'(0))=1(3 \cdot 2+2 \cdot(-3))=1(6-6)=0$

So the equation of the tangent line is $y=4$

for the last one take the derivative of

$g'(x)=e^{-2x}(3f(x)+2f'(x))$

$g''(x)=(3f(x)+2f'(x))\frac{d}{dx}e^{-2x}+e^{-2x}\frac{d}{dx}(3f(x)+2f'(x))=$

$-2e^{-2x}(3f(x)+2f'(x))+e^{-2x}(3f'(x)+2f''(x))=$

$e^{-2x}(-6f(x)-4f'(x)+3f'(x)+2f''(x))$

$g''(x)=e^{-2x}(-6f(x)-f'(x)+2f''(x))$

I hope this helps.

3. 1. Consider the differential equation dy/dx = 3(x^2) / e^2y

(a) Find a solution for y = f(x) to the differential equation satisfying f(0) = ½.

(b) Find the domain and range of the function f found in part (a). Justify your answer.
The equation is seperable so

$\frac{dy}{dx}=\frac{3x^2}{e^{2y}} \iff e^{2y}dy=3x^2dx$

So we integrate both sides...

$\int e^{2y}dy= \int 3x^2dx \iff \frac{1}{2}e^{2y}=x^3+c$

Solving for y we get...

$e^{2y}=2x^3+C$ taking the ln

$2y = ln(2x^3+c) \iff y=\frac{1}{2}ln(2x^3+c)$

using the inital condition we get

$\frac{1}{2}=\frac{1}{2}ln(c) \therefore c= e$

so we get $y=\frac{1}{2}ln(2x^3+e)$

Good luck.

4. I'm sorry, i posted number 2 wrong. What you got was what it was supposed to be TheEmpty Set. Lol, thanks!