How do you use the limit definition to find the slope of the tangent line to the graph #3x^2-5x+2# at x=3?

Question

How do you use the limit definition to find the slope of the tangent line to the graph  #3x^2-5x+2# at x=3?

Isaiah Allen · Accepted Answer

Do a lot of algebra after applying the limit definition to find that the slope at #x=3# is #13#.

The limit definition of the derivative is: #f'(x)=lim_(h->0)(f(x+h)-f(x))/h# If we evaluate this limit for #3x^2-5x+2#, we will get an expression for the derivative of this function. The derivative is simply the slope of the tangent line at a point; so evaluating the derivative at #x=3# will give us the slope of the tangent line at #x=3#. With that said, let's get started: #f'(x)=lim_(h->0)(3(x+h)^2-5(x+h)+2-(3x^2-5x+2))/h# #f'(x)=lim_(h->0)(3(x^2+2hx+h^2)-5x-5h+2-3x^2+5x-2)/h# #f'(x)=lim_(h->0)(cancel(3x^2)+6hx+3h^2-cancel(5x)-5h+cancel(2)-cancel(3x^2)+cancel(5x)-cancel(2))/h# #f'(x)=lim_(h->0)(6hx+3h^2-5h)/h# #f'(x)=lim_(h->0)(cancel(h)(6x+3h-5))/cancel(h)# #f'(x)=lim_(h->0)6x+3h-5# Evaluating this limit at #h=0#, #f'(x)=6x+3(0)-5=6x-5# Now that we have the derivative, we just need to plug in #x=3# to find the slope of the tangent line there: #f'(3)=6(3)-5=18-5=13#

Aurora Alvarado · Answer

See the explanation section below if your teacher/textbook uses #lim_(xrarra)(f(x)-f(a))/(x-a)# Some presentations of calculus use, for the defintion of the slope of the line tangent to the graph of #f(x)# at the point where #x=a# is #lim_(xrarra)(f(x)-f(a))/(x-a)# provided that the limit exists. (For example James Stewart's 8th edition Calculus p 106. On page 107, he gives the equivalent #lim_(hrarr0)(f(a+h)-f(a))/h#.) With this definition, the slope of the tangent line to the graph of #f(x) = 3x^2-5x+2# at the point where #x=3# is #lim_(xrarr3)(f(x)-f(3))/(x-3) = lim_(xrarr3)([3x^2-5x+2]-[3(3)^2-5(3)+2])/(x-3)# # = lim_(xrarr3)(3x^2-5x+2-27+15-2)/(x-3)# # = lim_(xrarr3)(3x^2-5x-12)/(x-3)# Note that this limit has indeterminate form #0/0# because #3# is a zero of the polynomial in the numerator. Since #3# is a zero, we know that #x-3# is a factor. So we can factor, reduce and try to evaluate again. # = lim_(xrarr3)(cancel((x-3))(3x+4))/cancel((x-3))# # = lim_(xrarr3)(3x+4) = 3(3)+4 = 13#. The limit is #13#, so the slope of the tangent line at #x=3# is #13#.

William Abdalla · Answer

undefined To find the slope of the tangent line to the graph of the function 3x^2-5x+2 at x=3 using the limit definition, we can follow these steps:

1. Start with the given function: f(x) = 3x^2-5x+2.

2. Determine the derivative of the function f(x) using the limit definition of the derivative. The derivative, denoted as f'(x), represents the slope of the tangent line at any given point.

3. Apply the limit definition of the derivative by taking the limit as h approaches 0 of the difference quotient: f'(x) = lim(h→0) [f(x+h) - f(x)] / h.

4. Substitute the given value of x=3 into the derivative expression: f'(3) = lim(h→0) [f(3+h) - f(3)] / h.

5. Evaluate the expression by plugging in the value of x=3 into the original function: f(3) = 3(3)^2 - 5(3) + 2.

6. Simplify the expression: f(3) = 3(9) - 15 + 2 = 27 - 15 + 2 = 14.

7. Substitute the value of f(3) into the derivative expression: f'(3) = lim(h→0) [f(3+h) - 14] / h.

8. Expand the expression f(3+h) by substituting x=3+h into the original function: f(3+h) = 3(3+h)^2 - 5(3+h) + 2.

9. Simplify the expression: f(3+h) = 3(9+6h+h^2) - 15 - 5h + 2 = 27 + 18h + 3h^2 - 15 - 5h + 2.

10. Combine like terms: f(3+h) = 3h^2 + 13h + 14.

11. Substitute the simplified expression for f(3+h) back into the derivative expression: f'(3) = lim(h→0) [(3h^2 + 13h + 14) - 14] / h.

12. Simplify the expression: f'(3) = lim(h→0) (3h^2 + 13h) / h.

13. Cancel out the h in the numerator and denominator: f'(3) = lim(h→0) (3h + 13).

14. Evaluate the limit as h approaches 0: f'(3) = 3(0) + 13 = 13.

Therefore, the slope of the tangent line to the graph of the function 3x^2-5x+2 at x=3 is 13.