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2 years ago

Show all work to factor x4 – 17x2 + 16 completely.

Mathematics

2 answers:

denis23 [38]2 years ago

4 0

Answer:

(x - 4)(x + 4)(x - 1)(x + 1)

Step-by-step explanation:

Given

$x^{4}$ - 17x² + 16

Use the substitution u = x², then

u² - 17u + 16

Consider the factors of the constant term (+ 16) which sum to give the coefficient of the u- term (- 17)

The factors are - 16 and - 1, then

u² - 17u + 16

= (u - 16)(u - 1) ← replace u by x²

= (x² - 16)(x² - 1) ← both factors are difference of squares

= (x - 4)(x + 4)(x - 1)(x + 1)

Anastaziya [24]2 years ago

3 0

Answer: -18

Step-by-step explanation: x4-17x2+16= -18

i am not completely sure but here. i hope this helps god bless

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Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.

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a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is $g^{-1}(x) = e^{x}$

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So, the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is $g^{-1}(x) = e^{x}$

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y = ln(x)

Taking exponents of both sides, we have

$e^{y} = e^{lnx} \\e^{y} = x$

Replacing x with y, we have

$y = e^{x}$

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So, the inverse image of g(x) is $g^{-1}(x) = e^{x}$

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

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Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

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