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Derivative and Tangent Line to f(x)

Created by @nathanedwards
 at November 1st 2023, 7:07:17 pm.
AP_Calculus_AB-Questions
#derivative
#tangent-line
#ap-calculus-ab

AP Calculus AB Exam Question:

Let f be a function defined by f(x)=3x2−2x+1xf(x) = \dfrac{3x^2 - 2x + 1}{x} for x≠0x\neq 0 and f(0)=4f(0) = 4.

(a) Find the derivative of f(x)f(x) using the definition of the derivative.

(b) Determine the equation of the tangent line to the graph of f(x)f(x) at the point (2, 7).

Answer:

(a) To find the derivative of f(x)f(x) using the definition of the derivative, we need to evaluate the limit:

f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \lim_{{h\to 0}}\dfrac{f(x+h)-f(x)}{h}

Step 1: Substitute the given function and simplify the expression:

f′(x)=lim⁡h→0(3(x+h)2−2(x+h)+1x+h)−(3x2−2x+1x)hf'(x) = \lim_{{h\to 0}}\dfrac{\left(\dfrac{3(x+h)^2 - 2(x+h) + 1}{x+h}\right) - \left(\dfrac{3x^2 - 2x + 1}{x}\right)}{h}

Step 2: Multiply each term by appropriate fractions to get a common denominator:

f′(x)=lim⁡h→0(3x2−2x+1)(x)−(3(x+h)2−2(x+h)+1)(x)hx(x+h)f'(x) = \lim_{{h\to 0}}\dfrac{(3x^2 - 2x + 1)(x) - (3(x+h)^2 - 2(x+h) + 1)(x)}{hx(x+h)}

Step 3: Expand and simplify the expression further:

f′(x)=lim⁡h→03x3−2x2+x−3x3−6x2h−3xh+2x2+2xh−hhx(x+h)f'(x) = \lim_{{h\to 0}}\dfrac{3x^3 - 2x^2 + x - 3x^3 - 6x^2h - 3xh + 2x^2 + 2xh - h}{hx(x+h)}
f′(x)=lim⁡h→0−6x2h−3xh+2xh−hhx(x+h)f'(x) = \lim_{{h\to 0}}\dfrac{-6x^2h - 3xh + 2xh - h}{hx(x+h)}
f′(x)=lim⁡h→0−6x2h−hhx(x+h)f'(x) = \lim_{{h\to 0}}\dfrac{-6x^2h - h}{hx(x+h)}

Step 4: Factor out 'h' from the numerator:

f′(x)=lim⁡h→0h(−6x2−1)hx(x+h)f'(x) = \lim_{{h\to 0}}\dfrac{h(-6x^2 - 1)}{hx(x+h)}

Step 5: Cancel out 'h' from the numerator and denominator:

f′(x)=lim⁡h→0−6x2−1x(x+h)f'(x) = \lim_{{h\to 0}}\dfrac{-6x^2 - 1}{x(x+h)}

Step 6: Substitute h = 0 in the expression:

f′(x)=−6x2−1x(x)f'(x) = \dfrac{-6x^2 - 1}{x(x)}
f′(x)=−6x2−1x2f'(x) = \dfrac{-6x^2 - 1}{x^2}

Therefore, the derivative of f(x)f(x) is f′(x)=−6x2−1x2f'(x) = \dfrac{-6x^2 - 1}{x^2}.

(b) To determine the equation of the tangent line to the graph of f(x)f(x) at the point (2, 7), we need to find the slope of the tangent line using the derivative and then use the point-slope form of a line.

Step 1: Evaluate the derivative at x = 2:

f′(2)=−6(2)2−1(2)2=−24−14=−254f'(2) = \dfrac{-6(2)^2 - 1}{(2)^2} = \dfrac{-24 - 1}{4} = -\dfrac{25}{4}

The slope of the tangent line is −254-\dfrac{25}{4}.

Step 2: Use the point-slope form of a line with the point (2, 7):

y−y1=m(x−x1)y - y_1 = m(x - x_1)
y−7=−254(x−2)y - 7 = -\dfrac{25}{4}(x - 2)

Step 3: Simplify the equation:

y−7=−254x+252y - 7 = -\dfrac{25}{4}x + \dfrac{25}{2}
y=−254x+392y = -\dfrac{25}{4}x + \dfrac{39}{2}

Therefore, the equation of the tangent line to the graph of f(x)f(x) at the point (2, 7) is y=−254x+392y = -\dfrac{25}{4}x + \dfrac{39}{2}.

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