Question

Function: g(x) = 2x2 – 8

For x ≥ 0, the inverse function is f(x)= √
1
2
x + 4

For x ≤ 0, the inverse function is d(x)= – √
1
2
x + 4

Answer 1

Answer:

$q=0,\ r=2,\ s=3,\ t=-3$

Step-by-step explanation:

We must fill in the missing values in the table.

That is, we must find the values of q, r, s, t

Note that:

$q = d (-8)$   then we evaluate the function d at $x = -8$

$q =-\sqrt{\frac{1}{2}(-8)+4}$

$q =-\sqrt{-4+4}$

$q =0$

Note that:

$r = f(0)$   then we evaluate the function f at $x = 0$

$r =\sqrt{\frac{1}{2}(0)+4}$

$r =\sqrt{4}$

$r =2$

Note that:

$s = f(10)$   then we evaluate the function f at $x = 10$

$s =\sqrt{\frac{1}{2}(10)+4}$

$s =\sqrt{5+4}$

$s =3$

Note that:

$t = d(10)$   then we evaluate the function d at $x = 10$

$t =-\sqrt{\frac{1}{2}(10)+4}$

$t =-\sqrt{5+4}$

$t =-3$

finally:

$q=0,\ r=2,\ s=3,\ t=-3$

Answer 2

for the function $g(x) = 2x^2-8$, q = 0, r = 2, s = 3, t = -3

The given function is:

$g(x) = 2x^2-8$

The inverse of $g(x) = 2x^2-8$ for x ≥ 0 is $f(x) = \sqrt{\frac{1}{2}x+4 }$

The inverse of $g(x) = 2x^2-8$ for  x ≤ 0 is  $d(x) = -\sqrt{\frac{1}{2}x+4 }$

For x = -8

$d(-8) = -\sqrt{\frac{1}{2}(-8)+4 }\\\\d(-8) = 0$

q = d(-8)  =  0

r = f(0)

$f(x) = \sqrt{\frac{1}{2}x+4 }\\\\f(0) = \sqrt{\frac{1}{2}(0)+4 }\\\\f(0)= \sqrt{4} \\\\f(0) = 2$

r = 2

s = f(10)

$f(x) = \sqrt{\frac{1}{2}x+4 }\\\\f(10) = \sqrt{\frac{1}{2}(10)+4 }\\\\f(10)= \sqrt{9} \\\\f(10) = 3$

s  =  3

t  =  d(10)

$d(x) = -\sqrt{\frac{1}{2}x+4 }\\\\d(10) = -\sqrt{\frac{1}{2}(10)+4 }\\\\d(10)= -\sqrt{9} \\\\d(10) = -3$

Therefore, for the function $g(x) = 2x^2-8$, q = 0, r = 2, s = 3, t = -3

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