AP CALCULUS open response...?

Question

Let f be the function defined by f(x) = (x^2-3)e^x for all real numbers x.

a. For what value of x is f increasing?

b. Find the x-coordinate of each point of inflection of f.

c. Find the x- and y-coordinates of the point, if any, where f(x) attains its absolute! minimum.

Answer 1

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

0 = (2x)e^x + (x^2 - 3)e^x
0 = e^x(2x + x^2 - 3)

e^x cannot equal 0, but the other factor can:

x^2 + 2x - 3 = 0
(x-1)(x+3) = 0
x = 1,-3

f(x) increases from (-infinity, -3) U ( 1, infinity)


f'(x) = (2x)e^x + (x^2 - 3)e^x
f"(x) = 2e^x + (2x)e^x + (2x)e^x + (x^2 - 3)e^x
f"(x) = e(x) (2 + 2x + 2x + x^2 -3)
0 = e(x) (2 + 2x + 2x + x^2 -3)

x^2 +4x -1 = 0

x = [-4 + /- sqrt(16+4) ]/2
x = [-4 +/- sqrt(20) ]/2
x = [-4 +/- 2sqrt(5) ]/2

x = -2 +/- sqrt(5)

y = (x^2-3)e^x

B) (-2 + sqrt(5), -3.728215) or (-2 - sqrt(5), .21615924)



minimum = f'(x) changes from decreasing to increasing...

min at (1, -5.436564)

Answer 2

f '(x) = 2xe^x+ (x^2-3)e^x = e^x(x^2+2x-3)=e^x(x+3)(x-1)
The derivative is 0 at x = -3 and x = 1
x+ - 3 is where a local max occurs and x = 1 is where a local min occurs.

f is increasing when f ' is positive and decreasing when f ' is negative. You should be able to figurebthat out easily since e^x is always positive.

Take second derivative to find inflection points.

Answer 3

If the backside of the ladder is x ft faraway from the backside of the wall then the suitable might desire to be ?(15^2-x^2) ft away with the help of Pythagoras. Now the backside of the ladder is shifting a nil.5ft/sec so which you would be able to describe the placement of the backside as x = 0.5t after t seconds (we can in basic terms pretend it started vertical as we're not relatively fascinated in the place it started.) So at time t the suitable of the ladder would be top = ?(15^2-(0.5t)^2) = ?(225-.25t^2) you prefer to discover the value so we would desire to discover the spinoff of the top with admire to t. dh/dt = -.25t / ?(225-0.25t^2) while the backside is 9ft away t might desire to be 18 seconds to slot the above equation (x=0.5t) so we would desire to sub in t = 18 into dh/dt. dh/dt (at 18) = -.25*18 / ?(225-0.25*18^2) = -0.375. So the dimensions OY is changing at reducing at 0.375ft/sec while OX is 9. the component of the triangle is one million/2 base instances top which making use of the above length is: A = 0.5*x*h A = 0.25t*?(225-.25t^2) making use of the product rule: dA/dt = 0.25*?(225-.25t^2) + 0.25t*-.25t / ?(225-.25t^2) dA/dt = 0.25(225-.5t^2) / ?(225-.25t^2) Subbing in t = 18 lower back supplies: dA/dt (at 18) = 21/sixteen = one million.3125 squareft/sec word: the 2d answer says dx/dt=2 (given) while actually its 0.5. additionally Adam accidently dropped a 2 off the backside of his fraction going to the line 9/sqrt(a hundred and forty four) so his answer is two times what this is going to be i.e. -.375 and he then makes use of this answer for section b which messes it up too. in case you persist together with his steps and perfect this you get the comparable answer as me.