Question

Use the Mean Value Theorem to prove that if the derivative of f(x) is less than or equal to 2 for x>0 and f(0)=4 then f(x) is less than or equal to 2x+4 for all x greater than or equal to 0.

Answer 1

Answer:

The proof is given below.

Step-by-step explanation:

Given:

$f'(x)\leq 2,\ f(0)=4, \ x>0$

Using Mean Value Theorem,

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Here, $a=0,\ b=x$

Therefore, $f'(c)=\frac{f(x)-f(0)}{x-0}=\frac{f(x)-4}{x}$

Now, as per question the derivative of $f(x)$ is less than or equal to 2.

Therefore, $f'(c)\leq 2$. Plug in the value of $f'(c)$. This gives,

$\frac{f(x)-4}{x}\leq 2\\f(x)-4\leq 2x ...............(\textrm{Cross product})\\f(x)\leq 2x+4 ..............(\textrm{Adding 4 both sides})$

Therefore, we get, $f(x)\leq 2x+4$ and hence it is proved using mean value theorem.