f'(x) = -2(x+1)e^x
f"(x) = -2(x+2)e^x
(A) since f"(-1) < 0, f is concave downward -- has a maximum there
(B) increasing: x > -1
concave down: x < -2
so, both: never
(C) f(x) = -2xe^x + C
since f(0) = 3, C = -3 and
f(x) = -2xe^x + 3
f(-1) = 2/e + 3
I need HELP
4.The derivative of a function f is given by f'(x)=(-2x-2)e^x, and f(0) = 3.
A. The function f has a critical point at x = -1. At this point, does f have a relative minimum, a relative maximum, or neither? Justify your answer.
B. On what intervals, if any, is the graph of f both increasing and concave down? Explain your reasoning.
C. Find the value of f(-1).
4 answers
Hey thank you so much for the response. So is there a reason why you labeled the -2(x+2)e^x equation as a second derivative with f"
sure -- because that's what it is
f'(x) = (-2x-2)e^x = -2xe^x - 2e^x
f"(x) = -2e^x - 2xe^x - 2e^x = -2xe^x - 4e^x = -2(x+2)e^x
or are you asking why I used f" as the label?
Also because that's what's often used.
Then nth derivative for n > 2 is usually f(n)(x)
f'(x) = (-2x-2)e^x = -2xe^x - 2e^x
f"(x) = -2e^x - 2xe^x - 2e^x = -2xe^x - 4e^x = -2(x+2)e^x
or are you asking why I used f" as the label?
Also because that's what's often used.
Then nth derivative for n > 2 is usually f(n)(x)
Oh nevermind, that makes more sense, thank you though