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Question
- ★ multiple choice what is g(f(x)) if f(x) = 7x² and g(x) = 3x⁻²? a \\(\frac{3}{49x^4}\\) b 21 c 21x⁴ d \\(\frac{7}{9x^4}\\)
Step1: Substitute \( f(x) \) into \( g(x) \)
To find \( g(f(x)) \), we replace \( x \) in \( g(x) \) with \( f(x) = 7x^2 \). So we have \( g(f(x)) = g(7x^2) \).
Step2: Apply the function \( g \)
The function \( g(x) = 3x^{-2} \), so substituting \( x = 7x^2 \) into \( g(x) \), we get \( g(7x^2)=3(7x^2)^{-2} \).
Step3: Simplify the expression
Using the exponent rule \( (ab)^n=a^n b^n \) and \( (a^m)^n = a^{mn} \), we have:
Wait, this seems to be option A, but let's check again. Wait, maybe I made a mistake. Wait, \( g(x) = 3x^{-2}=\frac{3}{x^2} \). Then \( g(f(x))=g(7x^2)=\frac{3}{(7x^2)^2}=\frac{3}{49x^4} \). But wait, the options include B as 21. Wait, maybe I misread the function. Wait, is \( g(x)=3x - 2 \) or \( 3x^{-2} \)? The original problem says \( g(x)=3x^{-2} \). Wait, but let's re - evaluate. Wait, if \( f(x)=7x^2 \) and \( g(x)=3x^{-2} \), then \( g(f(x)) = 3(7x^2)^{-2}=3\times7^{-2}\times x^{-4}=\frac{3}{49x^4} \), which is option A. But maybe there is a typo in my understanding. Wait, if \( g(x)=3x - 2 \), then \( g(f(x))=3(7x^2)-2 = 21x^2-2 \), which is not an option. So according to the given functions, the correct simplification leads to option A. But wait, the user's options have B as 21. Wait, maybe the function \( g(x) \) is \( 3x - 2 \)? No, the original problem says \( g(x)=3x^{-2} \). Wait, let's check again. Wait, \( (7x^2)^{-2}=\frac{1}{(7x^2)^2}=\frac{1}{49x^4} \), then \( 3\times\frac{1}{49x^4}=\frac{3}{49x^4} \), which is option A. But let's check the options again. The options are A: \( \frac{3}{49x^4} \), B:21, C: \( 21x^4 \), D: \( \frac{7}{9x^4} \).
Wait, maybe I made a mistake in the exponent. Wait, \( (7x^2)^{-2}=7^{-2}x^{-4}=\frac{1}{49x^4} \), then multiplied by 3 gives \( \frac{3}{49x^4} \), which is option A.
But wait, maybe the function \( g(x) \) is \( 3x - 2 \)? No, the problem says \( 3x^{-2} \). So the correct answer should be option A.
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A. \(\frac{3}{49x^4}\)