I am not sure what to do here:
Find the derivative (f '(x) using the limit process
a). f(x)= 2x^2+5x+2 find f ' (2)
b.) Find the equation of the Tangent Line at (1,5)
suppressing the limit notation for the moment,
f' = [f(x+h)-f(x)]/h
f' = [(2(x+h)^2 + 5(x+h) + 2) - (2x^2 + 5x + 2)]/h
f' = [(2x^2 + 4xh + 2h^2) + (5x + 5h) + 2 - (2x^2 + 5x + 2)]/h
f' = (4xh + 2h^2 + 5h)/h
f' = 4x + 5 + 2h
lim as h->0 = 4x+5
-------------------------
f'(1) = 9
(y-5) = 9(x-1)
Thank you so so much! I finally understand what to do! :)
To find the derivative of a function using the limit process, you can follow these steps:
a). First, let's find the derivative of the function f(x) = 2x^2 + 5x + 2 using the limit process to determine f '(2).
Step 1: Start with the definition of the derivative:
f '(x) = lim(h→0) [(f(x + h) - f(x)) / h]
Step 2: Substitute x = 2 into the definition:
f '(2) = lim(h→0) [(f(2 + h) - f(2)) / h]
Step 3: Substitute the function f(x) = 2x^2 + 5x + 2 into the equation:
f '(2) = lim(h→0) [(2(2 + h)^2 + 5(2 + h) + 2 - (2(2)^2 + 5(2) + 2)) / h]
Step 4: Simplify the equation:
f '(2) = lim(h→0) [(2(4 + 4h + h^2) + 10 + 5h + 2 - (2(4) + 10 + 2)) / h]
= lim(h→0) [(2(4 + 4h + h^2) + 10 + 5h + 2 - 28 - 10 - 2) / h]
= lim(h→0) [(8 + 8h + 2h^2 + 10 + 5h + 2 - 28 - 10 - 2) / h]
= lim(h→0) [(2h^2 + 13h - 10) / h]
Step 5: Factor out an h from the numerator:
f '(2) = lim(h→0) [h(2h + 13 - 10) / h]
Step 6: Cancel out h in the numerator and denominator:
f '(2) = lim(h→0) (2h + 13 - 10)
Step 7: Evaluate the limit as h approaches zero:
f '(2) = 2(0) + 13 - 10
= 13 - 10
= 3
Therefore, the derivative of f(x) = 2x^2 + 5x + 2 evaluated at x = 2 is f '(2) = 3.
b). To find the equation of the tangent line at the point (1, 5), we need two pieces of information: the slope of the tangent line (which we can find from the derivative) and the point (1, 5).
From part a, we found that f '(x) = 2x + 5.
Step 1: Substitute x = 1 into the derivative f '(x) = 2x + 5:
f '(1) = 2(1) + 5
= 2 + 5
= 7
Step 2: We now have the slope of the tangent line, which is 7. Use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1), where (x1, y1) is the point (1, 5) and m is the slope.
Plugging in the values, we get:
y - 5 = 7(x - 1)
Step 3: Simplify the equation:
y - 5 = 7x - 7
Step 4: Move the constant term to the other side:
y = 7x - 7 + 5
= 7x - 2
Therefore, the equation of the tangent line at the point (1, 5) is y = 7x - 2.
To find the derivative using the limit process, you need to follow these steps:
Step 1: Write down the function f(x).
Step 2: Write down the formula for the derivative using the limit process:
f '(x) = lim(h->0) [f(x + h) - f(x)] / h
Step 3: Substitute the given value of x into the formula obtained in step 2.
Let's solve the two problems:
a) f(x) = 2x^2 + 5x + 2, find f '(2):
Step 1: Write down the function f(x) = 2x^2 + 5x + 2.
Step 2: Write down the formula for the derivative using the limit process:
f '(x) = lim(h->0) [f(x + h) - f(x)] / h
Step 3: Substitute x = 2 into the formula obtained in step 2:
f '(2) = lim(h->0) [f(2 + h) - f(2)] / h
Step 4: Simplify the expression:
f '(2) = lim(h->0) [2(2 + h)^2 + 5(2 + h) + 2 - (2^2 + 5(2) + 2)] / h
f '(2) = lim(h->0) [2(4 + 4h + h^2) + 10 + 5h + 2 - (4 + 10 + 2)] / h
f '(2) = lim(h->0) [8 + 8h + 2h^2 + 10 + 5h + 2 - 16 - 10 - 2] / h
f '(2) = lim(h->0) [2h^2 + 13h] / h
f '(2) = lim(h->0) 2h + 13
Step 5: Substitute h = 0 into the above expression:
f '(2) = 2(0) + 13
f '(2) = 13
Therefore, the derivative of f(x) = 2x^2 + 5x + 2 at x = 2 is f '(2) = 13.
b) To find the equation of the tangent line at (1, 5), you need to find the slope of the tangent line and the point where it touches the curve.
Step 1: Write down the given point: (1, 5).
Step 2: Find the derivative of f(x) (which represents the slope):
f '(x) = 4x + 5
Step 3: Substitute x = 1 into the derived function to find the slope at x = 1:
f '(1) = 4(1) + 5
f '(1) = 9
Therefore, the slope of the tangent line at (1, 5) is 9.
Step 4: Use the slope-intercept form of a line to write the equation of the tangent line:
y - y1 = m(x - x1)
y - 5 = 9(x - 1)
y - 5 = 9x - 9
y = 9x - 4
Therefore, the equation of the tangent line at (1, 5) is y = 9x - 4.