Question #96b06

3 Answers
Oct 19, 2016

See below.

Explanation:

#e^(-x)# is monotonic strictly decreasing for #x in [0, oo)# and
#x^3# is monotonic strictly increasing for #x in [0,1)# both functions are continuous so they must cross for a point #x_0# such that
#e^(-x_0) = x_0^3 < 1#. Note that for #x in [0,1)# we have
#x_0^3 < 1# and #e^(-x_0) < 1#

Oct 19, 2016

See answer below

Explanation:

You must use the intermediate value theorem
let #f(x)=e^-x-x^3# this is a continuous function on the interval #(0,1)#
#f(0)=e^0-0=1#
and #f(1)=e^-1-1<0#
Then there is a value #c ∈ (0,1) # such that # f(1 )< f(c) < f(0)#

Oct 26, 2016

It is explained below

Explanation:

Given #e^(-x)=x^3#, take natural log on both sides. It would be #-xln e=3ln x#
Or -x = 3 ln x. This expression signifies that for all real x, x cannot be less than or equal to 0.

Now, in the logarithmic equation the LHS has - x, while it has a positive expression 3ln x on the RHS. This would be possible only if x is less than 1

Thus it is proved that x would lie between 0 and 1