Find (g-f)(6) where f and g are the following functions:f(x)= x+7 and g(x)=x^2+2x-5

Mathematics

1 answer:

Kay [80]3 years ago

5 0

X^2+2x-5-(x+7)
x^2+x-12
plug in 6 for x
6^2+6-12
36+6-12=30

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Dominik [7]

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Liula [17]

Answer:

$\boxed{\boxed{x=\dfrac{\pi}{3}\ \vee\ x=\pi\ \vee\ x=\dfrac{5\pi}{3}}}$

Step-by-step explanation:

$\cos(3x)=-1\iff3x=\pi+2k\pi\qquad k\in\mathbb{Z}\\\\\text{divide both sides by 3}\\\\x=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\\\\x\in[0,\ 2\pi)$

$\text{for}\ k=0\to x=\dfrac{\pi}{3}+\dfrac{2(0)\pi}{3}=\dfrac{\pi}{3}+0=\boxed{\dfrac{\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=1\to x=\dfrac{\pi}{3}+\dfrac{2(1)\pi}{3}=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\dfrac{3\pi}{3}=\boxed{\pi}\in[0,\ 2\pi)\\\\\text{for}\ k=2\to x=\dfrac{\pi}{3}+\dfrac{2(2)\pi}{3}=\dfrac{\pi}{3}+\dfrac{4\pi}{3}=\boxed{\dfrac{5\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=3\to x=\dfrac{\pi}{3}+\dfrac{2(3)\pi}{3}=\dfrac{\pi}{3}+\dfrac{6\pi}{3}=\dfrac{7\pi}{3}\notin[0,\ 2\po)$