If f(x) and its inverse function, f^-1(x), are both plotted on the same coordinate plane, what is their point of intersection?

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

8 0

Answer:

(a, a)

Step-by-step explanation:

actually there are two cases, don't have intersection and have. if have intersection, then they intersect at line y = x or point (a, a) by definition of inverse function.

jek_recluse [69]3 years ago

7 0

Answer:

the answer is D (3,3)

boom baby

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How do i solve this problem

Vladimir79 [104]

Solve the following system:
{6 t - 5 s = -4 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{-2 r - 4 s - 4 t = -9 | (equation 3)

Swap equation 1 with equation 3:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{-r - 4 s + 3 t = -4 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Subtract 1/2 × (equation 1) from equation 2:
{-(2 r) - 4 s - 4 t = -9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 1 by -1:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 2 s + 5 t = 1/2 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Multiply equation 2 by 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 4 s + 10 t = 1 | (equation 2)
{0 r - 5 s + 6 t = -4 | (equation 3)
Swap equation 2 with equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r - 4 s + 10 t = 1 | (equation 3)
Subtract 4/5 × (equation 2) from equation 3:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+(26 t)/5 = 21/5 | (equation 3)

Multiply equation 3 by 5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+26 t = 21 | (equation 3)
Divide equation 3 by 26:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s + 6 t = -4 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 6 × (equation 3) from equation 2:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r - 5 s+0 t = (-115)/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 2 by -5:
{2 r + 4 s + 4 t = 9 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 2) from equation 1:
{2 r + 0 s+4 t = 25/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Subtract 4 × (equation 3) from equation 1:
{2 r+0 s+0 t = (-17)/13 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
{0 r+0 s+t = 21/26 | (equation 3)
Divide equation 1 by 2:
{r+0 s+0 t = (-17)/26 | (equation 1)
{0 r+s+0 t = 23/13 | (equation 2)
v0 r+0 s+t = 21/26 | (equation 3)
Collect results:Answer: {r = -17/26
{s = 23/13 {t = 21/26

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