Consider the function.
f (x) = 5x3 + 3x
(a) Compute the derivative at x = 3 using the limit definition.
f '(3) =
(b) Find an equation of the tangent line.
2
evaluate f(3)
evaluate (3+h) and simplify
then take
Limit [ (f(3+h) - f(3) ]/h as h --->0
You should be able to factor out an h from the top which would then cancel the h in the bottom
you should get
f'(3) = 138
b) when x=3, y = 138 (that's a coincidence)
so you have
slope of tangent is 138 and (3,138) is a point on it
y = 138x + b , plug in the point
138 = 138(3) + b
b = -276
so y = 138x - 276 is the tangent equation
To compute the derivative of the function f(x) at x = 3 using the limit definition, we follow these steps:
(a) Compute the derivative at x = 3 using the limit definition:
Step 1: Write down the limit definition of the derivative:
f '(x) = lim(h→0) [f(x + h) - f(x)] / h
Step 2: Substitute the given value of x = 3 into the limit definition:
f '(3) = lim(h→0) [f(3 + h) - f(3)] / h
Step 3: Substitute the function f(x) into the equation:
f '(3) = lim(h→0) [(5(3 + h)^3 + 3(3 + h)) - (5(3)^3 + 3(3))] / h
Step 4: Simplify the expression:
f '(3) = lim(h→0) [(5(27 + 9h + h^2) + 9 + 3h) - (135 + 9)] / h
= lim(h→0) [(135 + 45h + 5h^2 + 9 + 3h) - 144] / h
= lim(h→0) [5h^2 + 48h] / h
= lim(h→0) 5h + 48
= 48
Therefore, the derivative of the function f(x) at x = 3 is f '(3) = 48.
(b) To find the equation of the tangent line at x = 3, we can use the point-slope form of a line, which is given by:
y - y₁ = m(x - x₁)
where (x₁, y₁) is a point on the line, and m is the slope of the line. In this case, we know that the point (3, f(3)) lies on the tangent line, and the slope of the tangent line is equal to the derivative at x = 3, which is 48. Therefore, the equation of the tangent line is:
y - f(3) = 48(x - 3)
You can simplify this equation further by substituting the value of f(3) from the original function f(x).