i need to solve this : -3^x +4^x +5^x

Tuesday, December 1, 2015

i need to solve this : -3^x +4^x +5^x

Therefore, we'll have to prove that $\displaystyle{{4}}^{{x}}-{{3}}^{{x}}\lt{{6}}^{{x}}-$
5^x

We'll use Lagrange's theorem to prove the
inequality.

We'll choose a function f(x)= $\displaystyle{{t}}^{{x}}$ , such as,
if we'll differentiate it with respect to t, we'll get: $\displaystyle{f{'}}{\left({x}\right)}=$
x*t^(x-1).

According to Lagrange's theorem, applied over
an interval [a,b],we'll get:

f(b) - f(a) =
f'(c)(b-a)

We'll choose the intervals [3;4] and [5;6], such
as:

$\displaystyle{{6}}^{{x}}-{{5}}^{{x}}=$
x*c^(x-1)(6-5)

$\displaystyle{{6}}^{{x}}-{{5}}^{{x}}=$
x*c^(x-1)

$\displaystyle{{4}}^{{x}}-{{3}}^{{x}}=$
x*d^(x-1)(4-3)

$\displaystyle{{4}}^{{x}}-{{3}}^{{x}}=$
x*d^(x-1)

We'll have to prove that $\displaystyle{{4}}^{{x}}-{{3}}^{{x}}\lt{{6}}^{{x}}-{{5}}^{{x}}$
=gt x*d^(x-1) ltx*c^(x-1)

We'll reduce by x which is
positive and it will keep the direction of the inequality
unchanged:

$\displaystyle{{d}}^{{{x}-{1}}}\lt$
c^(x-1)

Since x is positive and d < c (d is in the
interval [3;4] and c is in the interval [5;6]) => $\displaystyle{{d}}^{{{x}-{1}}}$
< $\displaystyle{{c}}^{{{x}-{1}}}$

According to Lagrange's
theorem, the given inequality 4^x- $\displaystyle{{3}}^{{x}}\lt{{6}}^{{x}}-{{5}}^{{x}}$ is
verified.

Help Notes

Tuesday, December 1, 2015