"find the indicated function values"

Mathematics

1 answer:

igomit [66]3 years ago

4 0

A) f(-10)

meaning x = -10

the first subfunction applies only below that, namely x < -10.

the second subfunction does contain -10 in it since -10 ⩽ x ⩽ 10, so we use that one then,

x = -10, thus x² is (-10)², or 100.

b) f(10)

x = 10.

that will also be the second subfunction, since it's the one that contains 10, thus,

x = 10, x² is (10)² or 100.

c) f(11)

x = 11.

well, is not the second subfunction, since its domain goes up to 10, x⩽10.

looking at the third subfunction, that includes 11, since x > 10, thus,

x = 11,

(11)² + 10

121 + 10

131.

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Answer:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Step-by-step explanation:

<u>Given functions</u>:

$p(x)=2x-4$

$r(x)=\dfrac{6x-1}{9x+1}$

Solve for p(x) = r(x):

$\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}$

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

<u>Quadratic Formula</u>

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=18, \quad b=-40, \quad c=-3$

Substitute the values of a, b and c into the <u>quadratic formula</u> and solve for x:

$\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}$