Question #143762
Find all points (x, y) where the functions f(x), g(x), h(x) have the same value:
f(x) = 2^x−5 + 3, g(x) = 2x − 5, h(x) = 8/x+ 10
1
Expert's answer
2020-11-16T08:18:43-0500

f(x)=2x5+3,  g(x)=2x5,  h(x)=8/x+10f(x)=2^{x-5}+3,\ \ g(x)=2x-5, \ \ h(x)=8/x+10 .

Let us find all points (x,y)(x,y) where functions g(x)g(x) and h(x)h(x) have the same value.

So, g(x)=h(x)=yg(x)=h(x)=y .

2x5=8/x+102x-5=8/x+10 (multiply both sides of equation by xx )

2x25x=8+10x2x^2-5x=8+10x

2x215x8=02x^2-15x-8=0

x1,2=15±1524×2×(8)2×2=15±174x_{1,2}=\frac{15\pm \sqrt{15^2-4\times 2\times (-8)}}{2\times 2}=\frac{15\pm17}{4}

x1=8,  x2=1/2x_1=8,\ \ x_2=-1/2

Now we calculate y1=g(x1)=11y_1=g(x_1)=11 , y2=g(x2)=6y_2=g(x_2)=-6 .

We have two points (8, 11)(8,\ 11) and (1/2, 6)(-1/2,\ -6) where functions g(x)g(x) and h(x)h(x) have the same value.

Let is find which of these points satisfy the condition, that functions f(x)f(x) , g(x)g(x) and h(x)h(x) have the same value.

f(x1)=23+3=11=y1f(x_1)=2^3+3=11=y_1

f(x2)=21/25+3=1322+3=6=y2f(x_2)=2^{-1/2-5}+3=\frac{1}{32\sqrt{2}}+3\cancel{=}-6=y_2 .

Answer: (8, 11)(8,\ 11) .


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