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How do you solve this function : g[f(x)] if f(x) = 4x + 1 and #g(x) = 2x^2 - 5#?

Answer 1

To solve the function ( g[f(x)] ) given ( f(x) = 4x + 1 ) and ( g(x) = 2x^2 - 5 ), you substitute the expression for ( f(x) ) into the function ( g(x) ).

( g[f(x)] = g(4x + 1) )

Then, you evaluate ( g(4x + 1) ) by substituting ( 4x + 1 ) into the function ( g(x) ).

( g(4x + 1) = 2(4x + 1)^2 - 5 )

Now, you simplify the expression.

( g(4x + 1) = 2(16x^2 + 8x + 1) - 5 )

( g(4x + 1) = 32x^2 + 16x + 2 - 5 )

( g(4x + 1) = 32x^2 + 16x - 3 )

So, the solution to the function ( g[f(x)] ) is ( 32x^2 + 16x - 3 ).

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Answer 2

Chloe Alexander

#g(f(x)) = 32x^2+16x-3#

If #color(white)("XXX")color(blue)(f(x))=color(blue)(4x+1)# and #color(white)("XXX")g(color(red)(x)) = 2color(red)(x)^2-5#

Then #color(white)("XXX")g(color(blue)(f(x)))= 2(color(blue)(f(x)))^2-5#

#color(white)("XXXXXXX")=2(color(blue)(4x+1))^2-5#

#color(white)("XXXXXXX")=2(color(blue)(16x^2+8x+1))-5#

#color(white)("XXXXXXX")=32x^2+16x-3#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.