How do you use the limit definition to find the derivative of #f(x)=x^2-15x+7#?

Question

Isabella Ackerman · Accepted Answer

Expand, reduce and evaluate the limit.

I'll use #f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h# Long version explanation For #f(x) = x^2-15x+7#, notice that #f(x+h) = (x+h)^2-15(x+h)+7#. #f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h# # = lim_(hrarr0)(((x+h)^2-15(x+h)+7)-(x^2-15x+7))/h# Notice that, if we try to evaluate by substitutiton, we get the indeterminate form #0/0#. Expand the numerator: (note #(x+h)^2 = x^2+2xh+h^2# #" "# use FOIL if you need to) # = lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7)-(x^2-15x+7))/h# # = lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7-x^2+15x-7))/h# Now, some of the terms in the numerator add to #0# # = lim_(hrarr0)((color(red)(x^2)+2xh+h^2 color(green)(-15x) -15hcolor(blue)(+7) color(red)(-x^2) color(green)(+15x)color(blue)(-7)))/h# # = lim_(hrarr0)(2xh+h^2-15h)/h# We still get #0/0#, but we can factor and reduce # = lim_(hrarr0)(cancel(h)(2x+h-15))/cancel(h)_1# # = lim_(hrarr0)(2x+h-15)# And now we can evaluate the limit # = 2x+(0)-15 = 2x-15# So, #f'(x) = 2x-15# Short version #f'(x) = lim_(hrarr0)(f(x+h)-f(x))/h# # = lim_(hrarr0)(((x+h)^2-15(x+h)+7)-(x^2-15x+7))/h# # = lim_(hrarr0)((x^2+2xh+h^2-15x-15h+7-x^2+15x-7))/h# # = lim_(hrarr0)(2xh+h^2-15h)/h# # = lim_(hrarr0)(2x+h-15)# # = 2x-15#

Luke Alvarez · Answer

undefined To find the derivative of $ f(x) = x^2 - 15x + 7 $ using the limit definition, follow these steps:

1. Begin with the definition of the derivative: $ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $.

2. Substitute the function $ f(x) = x^2 - 15x + 7 $ into the definition.

3. Expand $ f(x+h) $ and $ f(x) $.

4. Simplify the expression by combining like terms.

5. Take the limit as $ h $ approaches 0.

6. Evaluate the limit to find the derivative $ f'(x) $.