For the function, find the point(s) on the graph at which the tangent line has slope 3. y=1/3 x^3-3x^2 +11x+15

moiraudjpdn

moiraudjpdn

Answered question

2022-08-07

For the function, find the point(s) on the graph at which the tangent line has slope 3.
y = 1 3 x 3 3 x 2 + 11 x + 15

Answer & Explanation

Riya Cline

Riya Cline

Beginner2022-08-08Added 17 answers

y = x 3 3 3 x 2 + 11 x + 15
d y d x = x 2 6 x + 11
Given, slope of tangent, i.e., d y d x = 3
x 2 6 x + 11 = 3
x 2 6 x + 8 = 0
( x 4 ) ( x 2 ) = 0
x = 2 , 4
Thus, corresponding values of y are:
(i) for x = 2
y = 2 3 3 3 × 2 2 + 11 × 2 + 15 = 83 3
(ii) for x = 4
y = 97 3
So, required points are (2,83/3) and (4,97/3).
Cheyanne Jefferson

Cheyanne Jefferson

Beginner2022-08-09Added 2 answers

y = x 2 6 x + 11
let slop = y = x 2 6 x + 11 = 3
x 2 6 x + 8 = 0
( x 2 ) ( x 4 ) = 0
x=2, 4
when x=2
y = ( 1 / 3 ) 2 3 3 2 2 + 11 2 + 15 = 83 / 3
x=4
y = 1 / 3 4 3 3 4 2 + 11 4 + 15 = 97 / 3
so the points which the slop is 3 are (2,83/3) and(4,97/3)

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