Consider the function f ( x ) = x 2 e x . (a) Find the points, if any, at which
ID: 2858709 • Letter: C
Question
Consider the function f(x) = x2ex.
(a) Find the points, if any, at which the graph of the function f has a horizontal tangent line. (If an answer does not exist, enter DNE.)
(b) Find an equation for each horizontal tangent line. (Enter your answers as a comma-separated list of equations. If an answer does not exist, enter DNE.)
(c) Solve the inequality f'(x) > 0. (Enter your answer using interval notation. If an answer does not exist, enter DNE.)
(d) Solve the inequality f'(x) < 0. (Enter your answer using interval notation. If an answer does not exist, enter DNE.)
Explanation / Answer
f(x) = x2ex
f '(x) = 2xex + x2ex ; since (uv)' = u'v + uv'
==> f '(x) = ex(2x + x2)
slope is horizontal ==> f '(x) = 0
==> ex(2x + x2) = 0
==> exx(x + 2) = 0
ex can never be equal to zero
==> x = 0 , x = -2
x = 0==> f(x) = 02e0 = 0
x = -2 ==> f(x) = (-2)2e-2 = 4/e2 = 0.54
Hence f(x) has horizontal tangent at (-2 , 0.54) and (0 , 0)
b) equation of horizontal tangents
at x = -2 the equation is y = 0.54
at x = 0 the equation is y = 0
c) f '(x) > 0
==> exx(x + 2) > 0
ex ia always greater than zero
==> x(x +2) > 0
==> x < -2 , x > 0
Hence f '(x) > 0 ==> (- , -2) U (0 , )
d) f '(x) < 0
==> exx(x + 2) < 0
==> x(x + 2) < 0
==> -2 < x < 0
==> f '(x) < 0 ==> (-2 , 0)