f ( x ) = 2 x − 5 + 3 , g ( x ) = 2 x − 5 , h ( x ) = 8 / x + 10 f(x)=2^{x-5}+3,\ \ g(x)=2x-5, \ \ h(x)=8/x+10 f ( x ) = 2 x − 5 + 3 , g ( x ) = 2 x − 5 , h ( x ) = 8/ x + 10 .
Let us find all points ( x , y ) (x,y) ( x , y ) where functions g ( x ) g(x) g ( x ) and h ( x ) h(x) h ( x ) have the same value.
So, g ( x ) = h ( x ) = y g(x)=h(x)=y g ( x ) = h ( x ) = y .
2 x − 5 = 8 / x + 10 2x-5=8/x+10 2 x − 5 = 8/ x + 10 (multiply both sides of equation by x x x )
2 x 2 − 5 x = 8 + 10 x 2x^2-5x=8+10x 2 x 2 − 5 x = 8 + 10 x
2 x 2 − 15 x − 8 = 0 2x^2-15x-8=0 2 x 2 − 15 x − 8 = 0
x 1 , 2 = 15 ± 1 5 2 − 4 × 2 × ( − 8 ) 2 × 2 = 15 ± 17 4 x_{1,2}=\frac{15\pm \sqrt{15^2-4\times 2\times (-8)}}{2\times 2}=\frac{15\pm17}{4} x 1 , 2 = 2 × 2 15 ± 1 5 2 − 4 × 2 × ( − 8 ) = 4 15 ± 17
x 1 = 8 , x 2 = − 1 / 2 x_1=8,\ \ x_2=-1/2 x 1 = 8 , x 2 = − 1/2
Now we calculate y 1 = g ( x 1 ) = 11 y_1=g(x_1)=11 y 1 = g ( x 1 ) = 11 , y 2 = g ( x 2 ) = − 6 y_2=g(x_2)=-6 y 2 = g ( x 2 ) = − 6 .
We have two points ( 8 , 11 ) (8,\ 11) ( 8 , 11 ) and ( − 1 / 2 , − 6 ) (-1/2,\ -6) ( − 1/2 , − 6 ) where functions g ( x ) g(x) g ( x ) and h ( x ) h(x) h ( x ) have the same value.
Let is find which of these points satisfy the condition, that functions f ( x ) f(x) f ( x ) , g ( x ) g(x) g ( x ) and h ( x ) h(x) h ( x ) have the same value.
f ( x 1 ) = 2 3 + 3 = 11 = y 1 f(x_1)=2^3+3=11=y_1 f ( x 1 ) = 2 3 + 3 = 11 = y 1
f ( x 2 ) = 2 − 1 / 2 − 5 + 3 = 1 32 2 + 3 = − 6 = y 2 f(x_2)=2^{-1/2-5}+3=\frac{1}{32\sqrt{2}}+3\cancel{=}-6=y_2 f ( x 2 ) = 2 − 1/2 − 5 + 3 = 32 2 1 + 3 = − 6 = y 2 .
Answer: ( 8 , 11 ) (8,\ 11) ( 8 , 11 ) .
Comments