Question 1160763
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Let's find f(x+h)


{{{f(x) = 4x^2 + 3x}}}


{{{f(x+h) = 4(x+h)^2 + 3(x+h)}}} Replace every x with (x+h)


{{{f(x+h) = 4(x^2+2xh+h^2) + 3(x+h)}}} FOIL rule


{{{f(x+h) = 4x^2+8xh+4h^2 + 3x+3h}}} Distribute. This is one way to write the result for step 1.


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Now subtract off f(x)


{{{f(x+h) - f(x) = 4x^2+8xh+4h^2 + 3x+3h - ( f(x) )}}}


{{{f(x+h) - f(x) = 4x^2+8xh+4h^2 + 3x+3h - ( 4x^2+3x )}}}


{{{f(x+h) - f(x) = 4x^2+8xh+4h^2 + 3x+3h -  4x^2 - 3x }}}


{{{f(x+h) - f(x) = (4x^2 -  4x^2)+8xh+4h^2 + (3x-3x)+3h}}}


{{{f(x+h) - f(x) = 8xh+4h^2+3h}}} This is the result for step 2.


Notice how each term has an 'h' in it


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That common h will be factored out to help cancel and simplify things as shown below


{{{(f(x+h) - f(x))/h = (8xh+4h^2+3h)/h}}}


{{{(f(x+h) - f(x))/h = (h(8x+4h+3))/h}}}


{{{(f(x+h) - f(x))/h = (highlight(h)(8x+4h+3))/highlight(h)}}}


{{{(f(x+h) - f(x))/h = (cross(h)(8x+4h+3))/cross(h)}}}


{{{(f(x+h) - f(x))/h = 8x+4h+3}}} The result for step 3.


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As h approaches infinity, the expression {{{8x+4h+3}}} will approach {{{8x+4*0+3}}}. All I did here was plug in h = 0. Afterward, you simplify to go from {{{8x+4*0+3}}} to {{{8x+3}}}


Therefore, if f(x) = 4x^2+3x, then f'(x) = 8x+3 is the derivative.


Side note: To find the slope of the tangent line at any given point (x,y), you would plug the x coordinate into f'(x) = 8x+3 and simplify.
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