Question

Is `f(x)=(1-e^x)/x^2` increasing or decreasing at `x=2`?

Answer 1

The function `f(x)` is decreasing at `x=2`

Explanation:

Calculate the first derivative and look at the sign

We need

`(u/v)'=(u'v-uv')/(v^2)`

Here,

`u(x)=1-e^x`, `=>`, `u'(x)=-e^x`

`v(x)=x^2`, `=>`, `v'(x)=2x`

Therefore,

`f'(x)=(-e^x*x^2-2x(1-e^x))/(x^4)`

`=(-xe^x-2+2e^x)/(x^3)`

When `x=2`

`f'(2)=(-2e^2-2+2e^2)/(8)=-1/4`

As `f'(2)<0`, the function `f(x)` is decreasing at `x=2`

graph{(y-(1-e^x)/(x^2))(y-1000(x-2))=0 [-4.93, 12.85, -4.124, 4.765]}

Answer 2

`f(x)" is decreasing at x =2"`

Explanation:

`"to determine if a function f(x) is increasing/decreasing at"`
`"x = a evaluate "f'(a)`

`• " if "f'(a)>0" then f(x) is increasing at x = a"`

`• " if "f'(a)<0" then f(x) is decreasing at x = a"`

`"differentiate "f(x)" using the "color(blue)"quotient rule"`

`"given "f(x)=(g(x))/(h(x))" then"`

`f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"`

`g(x)=1-e^xrArrg'(x)=-e^x`

`"h(x)=x^2rArrh'(x)=2x`

`rArrf'(x)=(-x^2e^x-2x(1-e^x))/(4x^2)`

`rArrf'(2)=(-4e^2-4+4e^2)/16=-1/4`

`"since "f'(2)<0" then f(x) is decreasing at x = 2"`
graph{(1-e^x)/(x^2) [-10, 10, -5, 5]}