Question 16
Last updated at Dec. 21, 2017 by Teachoo

Transcript

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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