SOLUTION: Let {{{ f(x) = x^3 + 11 }}} and {{{ a = 3 }}} Find and simplify the quotient: {{{ (f(x)-f(a))/(x - a) }}} = Then find the slope {{{ M[a] }}} of the line tangent to th

Algebra ->  Equations -> SOLUTION: Let {{{ f(x) = x^3 + 11 }}} and {{{ a = 3 }}} Find and simplify the quotient: {{{ (f(x)-f(a))/(x - a) }}} = Then find the slope {{{ M[a] }}} of the line tangent to th      Log On


   



Question 987875: Let +f%28x%29+=+x%5E3+%2B+11+ and +a+=+3+
Find and simplify the quotient:
+%28f%28x%29-f%28a%29%29%2F%28x+-+a%29+ =
Then find the slope +++M%5Ba%5D+++ of the line tangent to the graph of f at the point (a, f (a))

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Hello,
your post was written incorrectly from the Perl's point of view.
I corrected and saved it.
You can submit it again with my corrections.
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OK.  Now I am continuing to solve it.

f(x) - f(a) = %28x%5E3%2B11%29 - %28a%5E3%2B11%29 = x%5E3-a%5E3 = %28x-a%29%2A%28x%5E2+%2B+ax+%2B+a%5E2%29.

%28f%28x%29+-+f%28a%29%29%2F%28x-a%29 = %28%28x-a%29%2A%28x%5E2+%2B+ax+%2B+a%5E2%29%29%2F%28x-a%29 = %28x%5E2+%2B+ax+%2B+a%5E2%29.

It is the expression for for the slope of the tangent line to the curve  y = x%5E3+%2B+11  in vicinity of the point  (a,f(a).

If the point  x  tends to the point  a,  then the slope tends to  a%5E2+%2B+a%5E2+%2B+a%5E2 = 3a%5E2.

Thus   M%5Ba%5D = 3a%5E2.