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Mathematics

1 answer:

kondaur [170]2 years ago

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Answer:your welcome

Step-by-step explanation:

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If f(x) =3x+5 and g(x) =x-1, then f[g(0) =?

Kisachek [45]

G(0)= 0-1= -1

f(-1)= 3(-1)+5= -3+5= 2

the answer to the question is 2

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3 years ago

For the function, f(x)=(9x+7)/(2x+4)

AVprozaik [17]

A. $f(x)=\frac{9x+7}{2x+4}\\f'(x)=\frac{(9)(2x+4)-(9x+7)(2)}{(2x+4)^2}\\f'(x)=\frac{(18x+36)-(18x+14)}{(2x+4)(2x+4)}\\f'(x)=\frac{22}{4x^2+16x+16}\\f'(x)=\frac{11}{2x^2+8x+8}\\\frac{11}{(2x+4)(x+2)}=0\\\frac{1}{(2x+4)(x+2)}=0\\x=-\infty,\infty$ - There are no critical points because the graph is neither continuous nor smooth. There is a discontinuity at x = 2.

B. $\frac{1}{(2x+4)(x+2)}=0\\x=-\infty,\infty$ - The absolute maximum is f(lim⇒-2_-) = infinity. The absolute minimum is f(lim⇒-2_+) = -infinity. This applies to the interval [-10, 7].

C. $f(x)=\frac{9x+7}{2x+4}\\f(0)=\frac{9(0)+7}{2(0)+4}\\f(0)=\frac{7}{4}\\f(0)=1.75\\f(5)=\frac{9(5)+7}{2(5)+4}\\f(5)=\frac{45+7}{10+4}\\f(5)=\frac{52}{14}\\f(5)=\frac{26}{7}\\f(5)=3.714$ - The absolute maximum is f(5) = 26/7 or 3.714. The absolute mimimum is f(0) = 1.75. This applies to the interval [0, 5]. Proof: graph f(x) at [0, 5] on a graph or graphing calculator.