Find the equation(s) of the tangent(s) to the curve y = (x 3 −1)(x − 2) at the points where the curve intersects the x – axis.

OR

Find the intervals in which the function f(x) = −3 log (1 + x) + 4 log (2 + x) − 4/(2 + x) is strictly increasing or strictly decreasing.

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Question 16 Find the equation(s) of the tangent(s) to the curve y = (x3 −1)(x − 2) at the points where the curve intersects the x –axis. We need to find equation of tangent to curve at which y cuts x-axis First, let us find the points at which y cuts x-axis Since y = 0 on points on x-axis Putting y = 0 in equation of y 0 = (x3 −1)(x − 2) (x3 −1)(x − 2) = 0 x3 −1 = 0 x3 = 1 x3 = 13 x = 1 x – 2 = 0 x = 2 So, x = 1, 2 Thus, points are (1, 0) and (2, 0) Now, to find equation of tangent, we find slope We know that Slope of tangent =𝑑𝑦/𝑑𝑥 = ((𝑥^3−1)(𝑥−2))^′ = (𝑥^3 (𝑥−2)−1(𝑥−2))^′ = (𝑥^4−2𝑥^3−𝑥+2)^′ = 〖4𝑥〗^3−2〖 ×3𝑥〗^2−1 = 〖4𝑥〗^3−〖 6𝑥〗^2−1 For point (1, 0) Slope of tangent = 〖4𝑥〗^3−〖 6𝑥〗^2−1 Putting x = 1 = 〖4(1)〗^3−〖 6(1)〗^2−1 = 4 – 6 – 1 = –3 Equation of tangent passing through (1, 0) with slope –3 is 𝑦 −𝑦_1= 𝑚 (𝑥 − 𝑥1) 𝑦 −0=−3 (𝑥 −1) 𝒚 =−𝟑𝒙+𝟑 For point (2, 0) Slope of tangent = 〖4𝑥〗^3−〖 6𝑥〗^2−1 Putting x = 2 = 〖4(2)〗^3−〖 6(2)〗^2−1 = 4 × 8 – 6 × 4 – 1 = 32 – 24 – 1 = 7 Equation of tangent passing through (2, 0) with slope 7 is 𝑦 −𝑦_1= 𝑚 (𝑥 − 𝑥1) 𝑦 −0=7 (𝑥 −2) 𝒚 =𝟕𝒙−𝟏𝟒 Question 16Theory We follow the following steps Step 1: Calculating f’(x) Step 2: Putting f’(x) = 0 Step 3: Drawing graphs and finding interval Step 4: Finding in which interval f’(x) > 0 or f’(x) < 0 Find the intervals in which the function f(x) = −3 log (1 + x) + 4 log (2 + x) − 4/(2 + 𝑥) is strictly increasing or strictly decreasing. Given f(x) = −3 log (1 + x) + 4 log (2 + x) − 4/(2 + 𝑥) Step 1: Calculating f’(𝑥) 𝑓^′ (𝑥) = ("−3 log (1 + x) + 4 log (2 + x) − " 4/(2 + 𝑥))^′ 𝑓^′ (𝑥) = ("−3 log (1 + x)" )^′ + ("4 log (2 + x) " )^′ – (4/(2 + 𝑥))^′ 𝑓^′ (𝑥) = (−3)/((1 + 𝑥)) + 4/(2 + 𝑥) – 4 ×−1 ×(2+𝑥)^(−2) 𝑓^′ (𝑥) = (−3)/((1 + 𝑥)) + 4/(2 + 𝑥) + 4/(2 + 𝑥)^2 𝑓^′ (𝑥) = (−3(2 + 𝑥)^2 + 4(2 + 𝑥)(1 + 𝑥) + 4(1 + 𝑥))/((1 + 𝑥) (2 + 𝑥)^2 ) 𝑓^′ (𝑥) = (−3(4 + 𝑥^2 + 4𝑥)+ 4(2(1 + 𝑥) + 𝑥(1 + 𝑥)) + 4 + 4𝑥)/((1 + 𝑥) (2 + 𝑥)^2 ) 𝑓^′ (𝑥) = (−12 − 3𝑥^2 − 12𝑥 + 4𝑥^2 + 12𝑥 + 8 + 4 + 4𝑥)/((1 + 𝑥) (2 + 𝑥)^2 ) Step 2: Putting 𝑓^′ (𝑥) = 0 4𝑥 – 3 = 0 4𝑥 = 3 𝑥 = 3/4 𝑓^′ (𝑥) = (𝑥^2 + 4𝑥)/((1 + 𝑥) (2 + 𝑥)^2 ) 𝑓^′ (𝑥) = (𝑥(𝑥 + 4))/((1 + 𝑥) (2 + 𝑥)^2 ) Step 2: Putting 𝑓^′ (𝑥) = 0 (𝑥(𝑥 + 4))/((1 + 𝑥) (2 + 𝑥)^2 )= 0 𝑥(𝑥 + 4)= 0 So, x = 0, x = –4 Step 3: Drawing graph Before drawing graph, first we find domain of function Given f(x) = −3 log (1 + x) + 4 log (2 + x) − 4/(2 + 𝑥) Now, log (1 + x) is possible if 1 + x > 0 i.e x > –1 log (2 + x) is possible if 2 + x > 0 i.e. x > –2 4/(2 + 𝑥) is possible if x + 2 ≠ 0, i.e. x ≠ – 2 So, domain will be which satisfies all 3 conditions ∴ Domain is x > –1 i.e. (–1, ∞) Since domain is (–1, ∞) x = –4 is not possible ∴ x = 0 is the only solution Now, making graph We make graph from –1 to infinity and plot point 0 Point 𝑥 = 0 divides the line into 2 disjoint intervals i.e. (–1, 0] & [0 ,∞)

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