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Sliva [168]
3 years ago
10

Evaluate the function f (x) = -4x + 9 at f (3).

Mathematics
2 answers:
sweet [91]3 years ago
4 0
Substituting 3 for x:
-4(3)+9
-12+9
-3
Umnica [9.8K]3 years ago
3 0
<span>Plug in f(x) = 3 to solve 

</span>f (x) = -4x + 9

3 = -4x + 9 

3-9=-4x 

-6=-4x 

-6/-4=x 

6/4=x 

3/2=x 

x=3/2
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PLEASE HELP!!!!<br> Which of these pair of functions are inverse functions?
Mamont248 [21]

Answer:

Option B and C are correct.

Step-by-step explanation:

Inverse function: If both the domain and the range are R for a function f(x), and if f(x) has an inverse g(x) then:

f(g(x)) = g(f(x)) = x for every x∈R.

Let f(x) = \frac{1}{2}(\ln(\frac{x}{2}) -1) and g(x) = 2e^{2x+1}

Use logarithmic rules:

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  • \ln a^b = b\ln a

then, by definition;

f(g(x)) = f(2e^{2x+1}) =\frac{1}{2}(\ln(\frac{2e^{2x+1}}{2})-1) = \frac{1}{2}(\ln(e^{2x+1}}){-1) = \frac{1}{2} (2x+1-1) =\frac{1}{2}(2x) = x

g(f(x)) = g(\frac{1}{2}(\ln(\frac{x}{2}) -1)) = 2e^{2({\frac{1}{2}(\ln(\frac{x}{2}) -1})+1 2e^{(\ln(\frac{x}{2}) -1+1}=2e^{\ln(\frac{x}{2})} =2\cdot \frac{x}{2} = x

Similarly;

for f(x) = \frac{4 \ln(x^2)}{e^2} and g(x) = e^{\frac{e^2 \cdot x}{8} }

then, by definition;

f(g(x)) = f(e^{\frac{e^2 \cdot x}{8}}) =\frac{4 \ln {(\frac{e^2 \cdot x}{8})^2}}{e^2} = \frac{8 \ln {(\frac{e^2 \cdot x}{8})}}{e^2} =\frac{8\frac{e^2\cdot x}{8} }{e^2}=\frac{8e^2 \cdot x}{8e^2}=x

Similarly,

g(f(x)) = x

Therefore, the only option B and C are correct. As the pairs of functions are inverse function.

3 0
4 years ago
If c is between a and b then Ac= CB
rodikova [14]
It only works if C is the midpoint of AB
5 0
4 years ago
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Anna007 [38]
Hi

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8/2 = 4
8/3 = 2.7
So the first option is incorrect.

8/4 = 2
8/2 = 4
The second option is correct.

The correct answer is 1/4 and 1/2

Hope this helps! :)
5 0
3 years ago
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7 0
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Galina-37 [17]
6/12=X/18
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Answer: NJ=9
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3 years ago
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