Calculus 1-Derivatives and the Shapes of Graphs, guaranteed 100% Pass

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Derivatives and the Shapes of Graphs



The Significance of the First Derivative

Increasing/Decreasing Test

a. If 𝑓’(𝑥) > 0 on an interval, then 𝑓(𝑥) is increasing on that interval

b. If 𝑓’(𝑥) < 0 on an interval, then 𝑓(𝑥) is decreasing on that interval.



To find where 𝒇′ (𝒙) > 0 or where 𝒇′ (𝒙) < 0 we first want to find where
𝒇′ (𝒙) = 𝟎 or where 𝒇′(𝒙) is undefined. We then “test” the sign of 𝒇′(𝒙) at points
in between the points where 𝒇′ (𝒙) = 𝟎 or where 𝒇′(𝒙) is undefined.



Ex. Find where the function 𝑓 (𝑥 ) = 𝑥 3 − 3𝑥 2 − 9𝑥 + 2 is increasing and where
it is decreasing.



First find where 𝑓 ′ (𝑥 ) = 0.

𝑓 ′ (𝑥 ) = 3𝑥 2 − 6𝑥 − 9 = 3(𝑥 2 − 2𝑥 − 3)
= 3(𝑥 − 3)(𝑥 + 1) = 0 ⟹ 𝑥 = 3, −1.
So 𝑓 ′ (𝑥 ) = 0 when 𝑥 = 3, −1.



Next test the sign of 𝑓 ′ (𝑥) for a single point in each of the intervals: 𝑥 < −1,

−1 < 𝑥 < 3, and 3 < 𝑥. 𝑓 ′ (𝑥) will have the same sign for every point in the
interval.

, 2


To test the sign of 𝑓 ′ (𝑥) for 𝑥 < −1, choose any point in that interval, for
example 𝑥 = −2, and find the sign of 𝑓 ′ (𝑥).

𝑓 ′ (−2) = 3(−2 − 3)(−2 + 1) = 3(−5)(−1) = 15 > 0.

So 𝑓 ′ (𝑥 ) > 0 for every point in 𝑥 < −1.



To test the sign of 𝑓 ′ (𝑥) for −1 < 𝑥 < 3, choose any point in that interval, for
example 𝑥 = 0, and find the sign of 𝑓 ′ (𝑥).

𝑓 ′ (0) = 3(0 − 3)(0 + 1) = 3(−3)(1) = −9 < 0.

So 𝑓 ′ (𝑥 ) < 0 for every point in −1 < 𝑥 < 3.



To test the sign of 𝑓 ′ (𝑥) for 3 < 𝑥, choose any point in that interval, for example
𝑥 = 4, and find the sign of 𝑓 ′ (𝑥).

𝑓 ′ (4) = 3(4 − 3)(4 + 1) = 3(1)(5) = 15 > 0.

So 𝑓 ′ (𝑥 ) > 0 for every point in 3 < 𝑥.



sign of 𝑓 ′ (𝑥) ______+_______|________−_________|_____+____

−1 3



So 𝑓(𝑥) is increasing when 𝑥 < −1 or 3 < 𝑥.

𝑓(𝑥) is decreasing when −1 < 𝑥 < 3.

, 3


Ex. Find where the function 𝑓 (𝑥 ) = 𝑥 3 − 3𝑥 2 + 4 is increasing and where it is
decreasing.



First find where 𝑓 ′ (𝑥 ) = 0.

𝑓 ′ (𝑥 ) = 3𝑥 2 − 6𝑥 = 3𝑥 (𝑥 − 2) = 0 ⟹ 𝑥 = 0, 2.
So 𝑓 ′ (𝑥 ) = 0 when 𝑥 = 0, 2.



Next test the sign of 𝑓 ′ (𝑥) for a single point in each of the intervals: 𝑥 < 0,

0 < 𝑥 < 2, and 2 < 𝑥. 𝑓 ′ (𝑥) will have the same sign for every point in the
interval.

To test the sign of 𝑓 ′ (𝑥) for 𝑥 < 0, choose any point in that interval, for example
𝑥 = −1, and find the sign of 𝑓 ′ (𝑥).

𝑓 ′ (−1) = 3(−1)(−1 − 2) = 9 > 0.

So 𝑓 ′ (𝑥 ) > 0 for every point in 𝑥 < 0.



To test the sign of 𝑓 ′ (𝑥) for 0 < 𝑥 < 2, choose any point in that interval, for
example 𝑥 = 1, and find the sign of 𝑓 ′ (𝑥).

𝑓 ′ (1) = 3(1)(1 − 2) = −3 < 0.

So 𝑓 ′ (𝑥 ) < 0 for every point in 0 < 𝑥 < 2.