Answer:
See explanation
Explanation:
We are given f(x)=ln(1+x)-x+(1/2)x^2.
We are first ask to differentiate this.
We will need chain rule for first term and power rule for all three terms.
f'(x)=(1+x)'/(1+x)-(1)+(1/2)×2x
f'(x)=(0+1)/(1+x)-(1)+x
f'(x)=1/(1+x)-(1)+x
We are then ask to prove if x is positive then f is positive.
I'm thinking they want us to use the derivative part in our answer.
Let's look at the critical numbers.
f' is undefined at x=-1 and it also makes f undefined.
Let's see if we can find when expression is 0.
1/(1+x)-(1)+x=0
Find common denominator:
1/(1+x)-(1+x)/(1+x)+x(1+x)/(1+x)=0
(1-1-x+x+x^2)/(1+x)=0
A fraction can only be zero when it's numerator is.
Simplify numerator equal 0:
x^2=0
This happens at x=0.
This means the expression,f, is increasing or decreasing after x=0. Let's found out what's happening there. f'(1)=1/(1+1)-(1)+1=1/2 which means after x=0, f is increasing since f'>0 after x=0.
So we should see increasing values of f when we up the value for x after 0.
Plugging in 0 gives: f(0)=ln(1+0)-0+(1/2)0^2=0.
So any value f, after this x=0, should be higher than 0 since f(0)=0 and f' told us f in increasing after x equals 0.