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# Let f(x) = 5x+3 and g(x)=x^2-2. What is g(f(-1))?

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Let f(x) = 5x+3 and g(x)=x^2-2. What is g(f(-1))?

and

If f(x)=16/(5+3x), what is the value of [f^-1(2)]^-2?

Thanks :D

WhichWitchIsWhich  Nov 4, 2017

### Best Answer

#1
+91051
+2

Let f(x) = 5x+3 and g(x)=x^2-2. What is g(f(-1))?

$$f(x)=5x+3\\ f(-1)=5*-1+3=-2\\~\\ g(x)=x^2-2\\ g(f(-1))=g(-2)=(-2)^2-2=2$$

and

If f(x)=16/(5+3x), what is the value of [f^-1(2)]^-2?

$$f(x)=\frac{16}{5+3x}\\ let\;\; y=\frac{16}{5+3x} \qquad x\ne-\frac{5}{3}\\ 5y+3yx=16\\ 3yx=16-5y\\ x=\frac{16-5y}{3y}\\ f^{-1}(x)=\frac{16-5x}{3x}\qquad x\ne0\\ f^{-1}(2)=\frac{16-5*2}{3*2}\\ f^{-1}(2)=\frac{6}{6}\\ f^{-1}(2)=1\\ [f^{-1}(2)]^{-2}=1\\$$

Here is a graph:

https://www.desmos.com/calculator/vc8o90j6i4

Melody  Nov 4, 2017
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### 1+0 Answers

#1
+91051
+2
Best Answer

Let f(x) = 5x+3 and g(x)=x^2-2. What is g(f(-1))?

$$f(x)=5x+3\\ f(-1)=5*-1+3=-2\\~\\ g(x)=x^2-2\\ g(f(-1))=g(-2)=(-2)^2-2=2$$

and

If f(x)=16/(5+3x), what is the value of [f^-1(2)]^-2?

$$f(x)=\frac{16}{5+3x}\\ let\;\; y=\frac{16}{5+3x} \qquad x\ne-\frac{5}{3}\\ 5y+3yx=16\\ 3yx=16-5y\\ x=\frac{16-5y}{3y}\\ f^{-1}(x)=\frac{16-5x}{3x}\qquad x\ne0\\ f^{-1}(2)=\frac{16-5*2}{3*2}\\ f^{-1}(2)=\frac{6}{6}\\ f^{-1}(2)=1\\ [f^{-1}(2)]^{-2}=1\\$$

Here is a graph:

https://www.desmos.com/calculator/vc8o90j6i4

Melody  Nov 4, 2017

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