document.write( "Question 968481: Please help,\r
\n" ); document.write( "\n" ); document.write( "Consider the function f(x) = x2 − 1. Find the equation of the tangent to the graph of f(x) at x = 3. [NOTE: when calculating f′(3), use first principles.] \n" ); document.write( "
Algebra.Com's Answer #591814 by rothauserc(4718)\"\" \"About 
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I assume the function f(x) = x2 − 1 is really
\n" ); document.write( "f(x) = x^2 - 1
\n" ); document.write( "we need the first derivative of this function to find the slope
\n" ); document.write( "using first principles,
\n" ); document.write( "f'(x) = lim h->0 (f(x+h) - f(x))/h
\n" ); document.write( "now let us find f(x+h)
\n" ); document.write( "f(x+h) = (x+h)^2 - 1 = x^2 + 2xh + h^2 - 1
\n" ); document.write( "f(x+h) - f(x) = x^2 + 2xh + h^2 - 1 -x^2 + 1 = 2xh + h^2
\n" ); document.write( "(f(x+h) - f(x))/h = 2x + h
\n" ); document.write( "as h->0 we can drop the h and limit becomes 2x
\n" ); document.write( "so f'(x) = 2x
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\n" ); document.write( "therefore slope at x = 3 is
\n" ); document.write( "f'(3) = 2*3 = 6
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\n" ); document.write( "using f(x) = x^2 - 1 to find y when x = 3
\n" ); document.write( "y = 3^2 -1 = 8
\n" ); document.write( "therefore our point is (3, 8), so far we have
\n" ); document.write( "y = 6x + b
\n" ); document.write( "now find b when x = 3 and y = 8
\n" ); document.write( "8 = 6*3 + b
\n" ); document.write( "b = -10
\n" ); document.write( "and the equation of the line tangent to the function at x = 3 is
\n" ); document.write( "y = 6x - 10\r
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