13. solve for x a. 5 5^x+1-5 14. Show that the period of f(x)=cos(x) is 2pi 15. Solve the following in equality a. 3x-6lt 2(3-x)+5 C. 2(5-x)geqslant -2x+5 b (x^2-5x+6)(x^2-4)geqslant 0 d. 3x^2+8x+5gt 0 16. Suppose (gcirc f)(x)=4x^2-9 and g(x)=x^2-6x find f(x) 17. Which of the following is a relations are function or not?how? a. R= (x,y):vert xvert =vert yvert b. R= (x,y):x=vert yvert C. R= (x,y)=vert xvert -vert yvert =9 18. Let f(x)=3x+4 and g(x)=6x^2+15x-11 then determine a. fcirc g (g(2) f +g 1.9 (g)/(f) gof 19. Let f(x)=4x+5 and g(x)=x^2+2 then find f+g, f.gf/g

13. To solve for x in the equation $$5^{x+1}-5=0$$ , we can rewrite it as $$5^{x+1}=5$$ . Taking the logarithm of both sides, we get $$(x+1)\log_5(5)=\log_5(5)$$ , which simplifies to $$x+1=1$$ . Therefore, $$x=0$$ .

14. To show that the period of $$f(x)=cos(x)$$ is $$2\pi$$ , we can use the fact that the cosine function repeats its values every $$2\pi$$ radians. This means that for any value of x, $$cos(x+2\pi)=cos(x)$$ . Therefore, the period of $$f(x)=cos(x)$$ is $$2\pi$$ .

15.
a. To solve the inequality $$3x-6<2(3-x)+5$$ , we can simplify it to $$3x-6<6-2x+5$$ , which further simplifies to $$5x<15$$ . Dividing both sides by 5, we get $$x<3$$ .

b. To solve the inequality $$(x^2-5x+6)(x^2-4)\geq0$$ , we can factor the expression to $$(x-2)(x-3)(x+2)(x-2)\geq0$$ . The critical points are $$x=-2$$ , $$x=2$$ , $$x=3$$ , and $$x=6$$ . Testing the intervals between these critical points, we find that the inequality is satisfied for $$x\leq-2$$ or $$x\geq3$$ .

c. To solve the inequality $$2(5-x)\geq-2x+5$$ , we can simplify it to $$10-2x\geq-2x+5$$ , which further simplifies to $$10\geq5$$ . This inequality is always true, so the solution is all real numbers.

d. To solve the inequality $$3x^2+8x+5>0$$ , we can use the quadratic formula to find the roots of the corresponding quadratic equation $$3x^2+8x+5=0$$ . The roots are $$x=-1$$ and $$x=-\frac{5}{3}$$ . Testing the intervals between these roots, we find that the inequality is satisfied for $$x<-\frac{5}{3}$$ or $$x>-1$$ .

16. Given $$(g\circ f)(x)=4x^2-9$$ and $$g(x)=x^2-6x$$ , we can find $$f(x)$$ by substituting $$g(x)$$ into $$(g\circ f)(x)$$ . This gives us $$f(x)^2-6f(x)=4x^2-9$$ . Solving this quadratic equation for $$f(x)$$ , we get $$f(x)=2x+3$$ or $$f(x)=-2x-3$$ .

17.
a. $$R=\{(x,y):|x|=|y|\}$$ is a relation, but it is not a function because for each x, there are two possible values of y (one positive and one negative).

b. $$R=\{(x,y):x=|y|\}$$ is a relation, and it is a function because for each x, there is only one possible value of y.

c. $$R=\{(x,y):|x|-|y|=9\}$$ is a relation, but it is not a function because for each x, there are two possible values of y (one positive and one negative).

18.
a. $$f\circ g(g(2))=f(g(2))=f(6(2)^2+15(2)-11)=f(31)=3(31)+4=97$$

b. $$f+g(1.9)=f(1.9)+g(1.9)=3(1.9)+4+6(1.9)^2+15(1.9)-11=7.7+4+13.86+28.5-11=62.16$$

c. $$\frac{g}{f}(gof)=\frac{g(gof)}{f(gof)}=\frac{g(f(gof))}{f(gof)}=\frac{g(f(6(2)^2+15(2)-11))}{f(6(2)^2+15(2)-11)}=\frac{g(f(31))}{f(31)}=\frac{g(97)}{f(97)}=\frac{6(97)^2+15(97)-11}{3(97)+4}=\frac{56406+1455-11}{295}=\frac{57050}{295}=193.22$$

19.
a. $f+g(x)=f(x)+g(x)=4x+5+x^