Question

Help with 5b please . thank you.​

Answer 1

Answer:

See explanation

Step-by-step 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.