Question

consider these functions f(x)=1/3x^2+4 g(x)=9x-12 what is the value of g(f(x))​

Answer 1

Given:

Functions are

$f(x)=\dfrac{1}{3}x^2+4$

$g(x)=9x-12$

To find:

The value of $g(f(x)).$.

Solution:

We have,

$g(f(x))=g(\dfrac{1}{3}x^2+4)$        $[\because f(x)=\dfrac{1}{3}x^2+4]$

$g(f(x))=9(\dfrac{1}{3}x^2+4)-12$        $[\because g(x)=9x-12]$

$g(f(x))=9(\dfrac{1}{3}x^2)+9(4)-12$

$g(f(x))=3x^2+36-12$

$g(f(x))=3x^2+24$

Therefore, the required function is $g(f(x))=3x^2+24$.

Answer 2

Following are the calculation to the given function:

Given:

$f(x)=\frac{1}{3}x^2+4 \\\\g(x)=9x-12$

To find:

$g(f(x))=?$

Solution:

$f(x)=\frac{1}{3}x^2+4 \\\\g(x)=9x-12$

$\to g(f(x))=g( \frac{1}{3} x^2+4)$

                $=9( \frac{1}{3} x^2+4)-12 \\\\=\frac{9}{3}x^2+36-12 \\\\=3x^2+24$

Therefore, the answer is "$24+3x^2$" .

Learn more about the g(f(x)):

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