Problem 4: Given the functions f(x)=x^(2)-4,g(x)=3x+1 and h(x)=1-2x . Evaluate a. g(-2)+h(3) f. g(f(x)) b. (f-g)(-1) g. (g@h)((1)/(2)) c. (f*g)(x) h. f(f(x)) d. ((h)/(f))(-2) i. ((g-f)/(2h))(0) e. (f@h)(x) j. (f@g@h)(1)

<p><br />a. Note that g(-2) = 3(-2) + 1 = -5, h(3) = 1 - 2(3) = -5. Hence the result g(-2)+h(3) is -5 - 5 = -10.<br /> <br />b. We obtain (f-g)(-1) by calculating x=-1 input in both functions f and g, then calculates difference between both. By substituting, we get f((-1)^2 - 4) - g((-3 +1)) = f(1-4) - g(-2) = f(-3) - g(-2) = f(-3)-(-2) =1.<br /><br />c. (f•g)(x) = [(x^2-4) • (3x+1)] can be simplified by processing each term separately, ending up with 3x^3+x-4.<br /><br />d. From \( \left(\frac{h}{f}\right)(-2) \), we substitute -2 in function h and f to get \( \frac{1--4}{4-4} \) when trying to simplify, we end up with \( \frac{5}{0} \) which is undefined.<br /><br />e. To find \( (f \circ h)(x) \) we should substitute h(x) in place of the variable x in the function f(x). This results in f(h(x))=f(1-2x)= (1-2x)^2-4 = 4x^2+4x+1-4 and where simplified we have 4x^2-4x-3.<br /><br />f. We get g[f(x)] = 3(x^2 - 4) + 1, which can then be simplified to g(x)= 3x^2-12 +1 and attempting to simplify the equation further results in g(x)=3x^2-11.<br /><br />g. Insert value into h(x), then get h(1/2) = 1 - 2(1/2) = 1 - 1 = 0. Insert it into g(x), so that g[o(i)], g[0] = 3*0 + 1 = 1.<br /> <br />h. For f(f(x)), we should replace x with \(f(x)=x^2 - 4\) in the pre-given function f(x). We get \(f(f(x))=(x^2 - 4)^2 -4 = x^4 -8x^2 + 12\).<br /> <br />i. \( \left(\frac{g-f}{2 h}\right)(0) = \left\{\left[3x+1 - (x^2 - 4)\right]/(2(1 - 2x))\right\}\)_x = 0.<br /><br />= \(-\frac{1}{2}\).<br /> <br />j. For \( (f \circ g \circ h)(1)\), first plug in 1 to h(x)→ g(’ h(1)) → f[’ g(h(1))]. Resulting → f[’ g(1- 2×1)], g(-1) → f(’ −2 -1) → f(-3) → (-3)^2 -4 → 9- 4= 5.<br />Results show that g(-2)+h(3) is 7, g(f(x)) is 9x^2+6x-3, (f.g)(x) is 3x^3 + x -4, and \( (f \circ h)(x) \) is4x^2+4x-3 just to name a few.<br />]</p>