Forgot the method (1 Viewer)

Run hard@thehsc · Mar 23, 2022

pls yelp

yanujw · Mar 23, 2022

Show that d/dx LHS > d/dx RHS for the domain given, and then prove the inequality true for the start of the domain by substitution (x=0).

vishnay · Mar 23, 2022

Run hard@thehsc said:
View attachment 35263
pls yelp

if u want to think abt it intuitively then:

$\text{Consider}\ f(x)=x-\ln(1+x)$
$\text{Then},\ f^{\prime}(x)=1-\frac{1}{1+x}$
$\text{Let} \ f^{\prime}(x)=0$
$1-\frac{1}{1+x}=0$
$\frac{1}{1+x}=1$
$1+x=1$
$x=0$
Therefore, turning point at x=0.
$f^{\prime\prime}(x)=\frac{1}{(x+1)^2}\geq0 \ \text{for all} \ x > 0$
Therefore minimum at $x=0$ -->function is increasing for all $x>0$ --> $x-ln(1+x)>0$ ---> $x>\ln(1+x)$

Run hard@thehsc · Mar 24, 2022

@vishnay and @yanujw thank you!

5uckerberg · Mar 24, 2022

So @Run hard@thehsc what @vishnay showed you are actually a very common way to solve for inequalities with regard to the first step moving everything to one side and making the other side 0.

@vishnay you can use \rightarrow instead of going ---> as shown here $\rightarrow$

vishnay · Mar 24, 2022

5uckerberg said:
@vishnay you can use \rightarrow instead of going ---> as shown here $\rightarrow$

i know

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Forgot the method (1 Viewer)

Run hard@thehsc

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yanujw

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vishnay

God

Run hard@thehsc

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5uckerberg

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vishnay

God

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