(a)g(x)=(x−1)(3x+2) ⟹ (3x+2)g(x)=x−1 ⟹ 3xg(x)+2g(x)−x+1=0 ⟹ 2g(x)+1=x−3xg(x) ⟹ x=1+2g(x)1−3g(x) ⟹ g−1(x)=1+2x1−3x;x=/ 1/3(b)f(x)=4+x2+1;R(f):f≥5 ⟹ f(x)−4=x2+1 ⟹ (f(x)−4)2=x2+1 ⟹ (f(x)−4)2−1=x2 ⟹ x=(f(x)−4)2−1f−1(x)=(x−4)2−1;x≥5(a)\\ g(x)=\dfrac{(x-1)}{(3x+2)} \implies (3x+2)g(x)=x-1\\ \implies3xg(x)+2g(x)-x+1=0 \implies 2g(x)+1=x-3xg(x) \\ \implies x=\dfrac{1+2g(x)}{1-3g(x)} \\ \\\implies\ g^{-1}(x)=\dfrac{1+2x}{1-3x}; x{=}\mathllap{/\,}1/3\\ \\ (b) \\ f(x)=4+\sqrt{\smash[b]{x^2+1}}; R(f):f\ge5 \implies f(x)-4=\sqrt{x^2+1}\\ \implies (f(x)-4)^2=x^2+1\\ \implies (f(x)-4)^2-1=x^2\\ \implies x=\sqrt{(f(x)-4)^2-1}\\ f^{-1}(x)=\sqrt{{(x-4)^2-1}}; x\ge5(a)g(x)=(3x+2)(x−1)⟹(3x+2)g(x)=x−1⟹3xg(x)+2g(x)−x+1=0⟹2g(x)+1=x−3xg(x)⟹x=1−3g(x)1+2g(x)⟹ g−1(x)=1−3x1+2x;x=/1/3(b)f(x)=4+x2+1;R(f):f≥5⟹f(x)−4=x2+1⟹(f(x)−4)2=x2+1⟹(f(x)−4)2−1=x2⟹x=(f(x)−4)2−1f−1(x)=(x−4)2−1;x≥5
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments