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Sergio039 [100]
3 years ago
13

Which function has an inverse function a) f(x)=|x-4|+1 (b) f(x)=25x^2+70x+49 (c) x^4 (d) x+3÷7

Mathematics
2 answers:
stich3 [128]3 years ago
5 0

Answer:

f(x)=\dfrac{x+3}{7} is f^{-1}(x)=7x-3

Step-by-step explanation:

Property of inverse function:

The function should be bijective function ( one-to-one and onto)

Option A) f(x)=|x-4|+1

It is absolute function. Not one-to-one function.

Domain: All real (-∞,∞)

Range: [1,∞)

False ( Inverse not possible )

Option B) f(x)=25x^2+70x+49

It is quadratic function. Not one-to-one function.

Domain: All real (-∞,∞)

Range: [0,∞)

False ( Inverse not possible )

Option C) f(x)=x^4

It is polynomial function with even degree. Not one-to-one function.

Domain: All real (-∞,∞)

Range: [0,∞)

False ( Inverse not possible )

Option D) f(x)=\dfrac{x+3}{7}

It is linear function. one-to-one and onto

Domain: All real (-∞,∞)

Range: All real (-∞,∞)

True ( Inverse possible )

Inverse of f(x)=\dfrac{x+3}{7} is f^{-1}(x)=7x-3

Hence, The function inverse function f(x)=\dfrac{x+3}{7} is f^{-1}(x)=7x-3

Bad White [126]3 years ago
3 0

Answer: Hello there!

A function only can have an inverse if the function is injective and surjective (and continuous):

Then we need to see; if f(x1) = f(x2) = y, and x1 is different from x2, then f(x) has not an inverse:

a) f(x) = Ix - 4I + 1

for example, f(0) = I-4I + 1 = 5

and f(8) = I8 -4I + 1 = 4 + 1 = 5

then f(x) does not have an inverse

b) f(x) = 25x^2 + 70x + 49

This is a cuadratic function, wich is graphed as a arc going up or down, wich means that there are two values of x that give the same value for f(x), then this function has not inverse. (this will be the case for all even powers)

c) f(x) = x^4

Again, an even power. But let's probe it:

f(1) = 1^4 = 1

f(-1) = (-1)^4 = 1

f(x) does not have an inverse:

d) f(x) = x + 3/7

Ok, here we have a linear equation, wich means that is injective and surjective.

The inverse of this function can be g(x) = x - 3/7

proof:

f(g(x)) = f( x - 3/7) = (x - 3/7) + 3/7 = x

then f and g are inverses of each other.

(if in this case f(x) = (x + 3)/7 = x/7 + 3/7 is also a linear equation, so it is injective and surjective (and continuous), wich implies that has an inverse)

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Answer:

a) Binomial distribution, with n=20 and p=0.10.

b) P(x>1) = 0.6082

c) P(3≤X≤5) = 0.3118

d) E(X) = 2

e) σ=1.34

Step-by-step explanation:

a) As we have a constant "defective" rate for each unit, and we take a random sample of fixed size, the appropiate distribution to model this variable X is the binomial distribution.

The parameters of the binomial distribution for X are n=20 and p=0.10.

X\sim B(0.10,20)

b) The probability of k defective surge protectors is calculated as:

P(x=k) = \binom{n}{k} p^{k}q^{n-k}

In this case, we want to know the probability that more than one unit is defective: P(x>1). This can be calculated as:

P(x>1)=1-(P(0)+P(1))\\\\\\P(x=0) = \binom{20}{0} p^{0}q^{20}=1*1*0.1216=0.1216\\\\P(x=1) = \binom{20}{1} p^{1}q^{19}=20*0.1*0.1351=0.2702\\\\\\ P(x>1)=1-(0.1216+0.2702)=1-0.3918=0.6082

c) We have to calculate the probability that the number of defective surge protectors is between three and five: P(3≤X≤5).

P(3\leq X\leq 5)=P(3)+P(4)+P(5)\\\\\\P(x=3) = \binom{20}{3} p^{3}q^{17}=1140*0.001*0.1668=0.1901\\\\P(x=4) = \binom{20}{4} p^{4}q^{16}=4845*0.0001*0.1853=0.0898\\\\P(x=5) = \binom{20}{5} p^{5}q^{15}=15504*0*0.2059=0.0319\\\\\\P(3\leq X\leq 5)=P(3)+P(4)+P(5)=0.1901+0.0898+0.0319=0.3118

d) The expected number of defective surge protectors can be calculated from the mean of the binomial distribution:

E(X)=\mu_B=np=20*0.10=2

e) The standard deviation of this binomial distribution is:

\sigma=\sqrt{np(1-p)}=\sqrt{20*0.1*0.9}=\sqrt{1.8}=1.34

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