Given f(x)=x², h(x)=x_9, p(x)=square root x_4÷x+4 and v(x)=5x_3. Find the composition of the functions.

A. V(f(x))
B. h(p(16))
C. V(p(x))
D. h(v(1))
E. h(f(x))
F. V(h(f(x))

1 answer

To find the compositions of the functions given the definitions:

  • \( f(x) = x^2 \)
  • \( h(x) = x + 9 \)
  • \( p(x) = \frac{\sqrt{x}}{x} + 4 \)
  • \( v(x) = 5x + 3 \)

Let's evaluate each composition step-by-step:

A. \( V(f(x)) \)

  1. First, find \( f(x) \): \[ f(x) = x^2 \]
  2. Next, substitute \( f(x) \) into \( v(x) \): \[ V(f(x)) = V(x^2) = 5(x^2) + 3 = 5x^2 + 3 \]

B. \( h(p(16)) \)

  1. First, calculate \( p(16) \): \[ p(16) = \frac{\sqrt{16}}{16} + 4 = \frac{4}{16} + 4 = \frac{1}{4} + 4 = \frac{1}{4} + \frac{16}{4} = \frac{17}{4} \]
  2. Now substitute \( p(16) \) into \( h(x) \): \[ h\left(\frac{17}{4}\right) = \frac{17}{4} + 9 = \frac{17}{4} + \frac{36}{4} = \frac{53}{4} \]

C. \( V(p(x)) \)

  1. Next, substitute \( p(x) \) into \( v(x) \): \[ V(p(x)) = V\left(\frac{\sqrt{x}}{x} + 4\right) = 5\left(\frac{\sqrt{x}}{x} + 4\right) + 3 \] \[ = 5\cdot \frac{\sqrt{x}}{x} + 20 + 3 = \frac{5\sqrt{x}}{x} + 23 \]

D. \( h(v(1)) \)

  1. First, calculate \( v(1) \): \[ v(1) = 5(1) + 3 = 5 + 3 = 8 \]
  2. Then substitute \( v(1) \) into \( h(x) \): \[ h(8) = 8 + 9 = 17 \]

E. \( h(f(x)) \)

  1. Substitute \( f(x) \) into \( h(x) \): \[ h(f(x)) = h(x^2) = x^2 + 9 \]

F. \( V(h(f(x))) \)

  1. We already found \( h(f(x)) = x^2 + 9 \).
  2. Now substitute this into \( v(x) \): \[ V(h(f(x))) = V(x^2 + 9) = 5(x^2 + 9) + 3 = 5x^2 + 45 + 3 = 5x^2 + 48 \]

Finally, summarizing the results:

  • A. \( V(f(x)) = 5x^2 + 3 \)
  • B. \( h(p(16)) = \frac{53}{4} \)
  • C. \( V(p(x)) = \frac{5\sqrt{x}}{x} + 23 \)
  • D. \( h(v(1)) = 17 \)
  • E. \( h(f(x)) = x^2 + 9 \)
  • F. \( V(h(f(x))) = 5x^2 + 48 \)