Question

Use the functions f (x )equals x squared minus 3 x minus 10 and g (x )equals x squared plus 3 x minus 10 to answer parts​ (a)-(g).​(a) Solve f (x )equals 0.​(d) Solve f (x )greater than 0.​(g) Solve f (x )greater than or equals 2.​(b) Solve g (x )equals 0.​(e) Solve g (x )less than or equals 0.​(c) Solve f (x )equals g (x ).​(f) Solve f (x )greater than g (x ).

Answer 1

Answer:

a)  x₁ = 5        x₂  = -2

b) x ∈ ( -∞ , -2 ) ∪  ( 5 , +∞ )

c ) x ∈ ( - ∞ . -3 ) ∪ ( 4 . + ∞ )

d)  x₁ = -5        x₂ = 2

e) x∈ [ -5 , 2 ]

f ) x = 0

g) x ∈ ( - ∞ , 0 )

Step-by-step explanation:

f(x) = x² - 3x -10

a) Solve  f(x) = 0

x² - 3x -10 = 0   ⇒    factoring we look for two numbers ( a and b ) in such way that

a*b  = - 10      and a + b = -3

The numbers are   -5    and 2, then we can arrange the equation as follows

( x - 5 ) * ( x + 2 )  = 0

Solving that we get

x - 5 = 0       x  =  5

x + 2 = 0       x  = -2

b) f(x) > 0

( x - 5 ) * ( x + 2 )  See annex  we graph roots of the equation and evaluate values of f(x) for points inside intervals, therefore

f(-3)  = (-3 - 5 ) * ( -3 + 1 )  = (-8) * (-2) = 16 and conclude, all values smaller than -2 are valids solution for f(x) > 0

f(0) = (0 - 5) * ( 0 + 2 ) = -10

f(0) < 0 then there is not solution between -2 and 5

f(6) = ( 6 - 5 ) * ( 6 + 2 ) = 8

f(6) = 8     f(6) > 0

So we conclude

x ∈ ( -∞ , -2 ) ∪  ( 5 , +∞ )

c) f(x) ≥ 2

x² - 3x -10 ≥ 2

x² - 3x -10 -2 ≥ 0

x² - 3x -12 ≥ 0

Factoring we get

( x - 4 ) * ( x + 3 ) ≥ 0

f(-4) = -8 * -1  = 8 ≥ 0   all values small than -3 are solutions

f(0) = -4 * 3  = -12  between -3  and 4 there are not solutions

f(5) = 1 * 8 = 8 ≥ 0

So we conclude

x ∈ ( - ∞ . -3 ) ∪ ( 4 . + ∞ )

g(x) = x² + 3x - 10

g(x) = 0

x² + 3x - 10  = 0

Factoring

( x + 5 ) * ( x - 2 ) = 0

x₁ = -5

x₂ = 2

g(x) ≤ 0

( x + 5 ) * ( x - 2 ) ≤ 0

g(-6)  = -1 * -8  = 8 ≥ 0     non solutions interval

g(0) = 5 * -2  = - 10  ≤ 0     solutions interval

g(3) = 8 * 1  = 8 ≥ 0   non solutions interval

We can see that the roots of the equation are valids solutions, then

x∈ [ -5 , 2 ]

f(x) = g(x)

x² - 3x -10 = x² + 3x - 10

- 6x = 0

x = 0

x² - 3x -10 >  x² + 3x - 10

Evaluating for negative values

( -1)² - 3(-1) - 10 = 1 +3 - 10 = -6     f(-1)

( -1)² +3(-1) - 10 = 1 -3 -10  = -12     g(-1)

f(-1) > g(-1)

Evaluating for positive values

f(1) = (1)² - 3 (1) - 10 = - 12

g(1) = (1)² + 3 (1) - 10 = - 6  

then g(1) > f(1)

f(0) = 10

g(0) = 10

Then f(x)  > g(x)   only for negative values or

f(x) > g(x)    x ∈ ( - ∞ , 0 )