Cho ham so f(x) xd tren R thoa man f'(x)=can e^x+e^{-x}-2, f(0)=5; f(in 1/4) =0 Tinh S=f(-In 16)+f(IN 4)

Đáp án:

$S=f(-\text{In}\;16)+f(\text{In}\; 4)=\dfrac 52$ 

Giải thích các bước giải:

 Ta có: $f'(x)=\sqrt{e^x+e^{-x}-2}=\dfrac{|e^x-1|}{\sqrt{e^x}}$

$+)$ $x<0\to f(x)=e^{-\dfrac x2}-e^{\dfrac x2}\to f(x)=-2e^{\dfrac {-x}2}-2e^{\dfrac x2}+C_2$

$+)$ $x\geqslant 0\to f(x)=e^{\dfrac x2}-e^{\dfrac {-x}2}\to f(x)=2e^{\dfrac x2}+2e^{-\dfrac x2}+C_1$

mà $f(0)=5\to 2e^0+2e^0+C_1=5\to C_1=1.$ Do đó, ta có:

$f(\text{In}\; 4)=1+2e^{\dfrac{\text{In}\;4}2}+2e^{-\dfrac{\text{In}\;4}2}=6$

Từ $f\left(\text{In}\;\dfrac 14\right)=0.$ Tương tự, ta có:

\(-2e^{-\dfrac{\text{In}\;\left(\dfrac 14\right)}2}-2e^{\dfrac{\text{In}\;\left(\dfrac 14\right)}2}=0\to 5=C_2\)

\(\to f(-\text{In}\;16)=-2e^{\dfrac{(-\text{In}16)}2}+\left(-2e^{\dfrac{-\text{In}\; 16}2}\right)+5\)

\(\to f(-\text{In}\;16)=-\dfrac 72\)

$\to S=f(-\text{In}\;16)+f(\text{In}\; 4)=\dfrac{-7}{2}+6=\dfrac 52$