<< Chapter < Page Chapter >> Page >
  • Explain how the sign of the first derivative affects the shape of a function’s graph.
  • State the first derivative test for critical points.
  • Use concavity and inflection points to explain how the sign of the second derivative affects the shape of a function’s graph.
  • Explain the concavity test for a function over an open interval.
  • Explain the relationship between a function and its first and second derivatives.
  • State the second derivative test for local extrema.

Earlier in this chapter we stated that if a function f has a local extremum at a point c , then c must be a critical point of f . However, a function is not guaranteed to have a local extremum at a critical point. For example, f ( x ) = x 3 has a critical point at x = 0 since f ( x ) = 3 x 2 is zero at x = 0 , but f does not have a local extremum at x = 0 . Using the results from the previous section, we are now able to determine whether a critical point of a function actually corresponds to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a graph by describing whether the graph of a function curves upward or curves downward.

The first derivative test

Corollary 3 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval I then the function is increasing over I . On the other hand, if the derivative of the function is negative over an interval I , then the function is decreasing over I as shown in the following figure.

This figure is broken into four figures labeled a, b, c, and d. Figure a shows a function increasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f’ > 0. In other words, f is increasing. Figure b shows a function increasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f’ > 0. In other words, f is increasing. Figure c shows a function decreasing concavely from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f’ < 0. In other words, f is decreasing. Figure d shows a function decreasing convexly from (a, f(a)) to (b, f(b)). At two points the derivative is taken and it is noted that at both f’ < 0. In other words, f is decreasing.
Both functions are increasing over the interval ( a , b ) . At each point x , the derivative f ( x ) > 0 . Both functions are decreasing over the interval ( a , b ) . At each point x , the derivative f ( x ) < 0 .

A continuous function f has a local maximum at point c if and only if f switches from increasing to decreasing at point c . Similarly, f has a local minimum at c if and only if f switches from decreasing to increasing at c . If f is a continuous function over an interval I containing c and differentiable over I , except possibly at c , the only way f can switch from increasing to decreasing (or vice versa) at point c is if f changes sign as x increases through c . If f is differentiable at c , the only way that f . can change sign as x increases through c is if f ( c ) = 0 . Therefore, for a function f that is continuous over an interval I containing c and differentiable over I , except possibly at c , the only way f can switch from increasing to decreasing (or vice versa) is if f ( c ) = 0 or f ( c ) is undefined. Consequently, to locate local extrema for a function f , we look for points c in the domain of f such that f ( c ) = 0 or f ( c ) is undefined. Recall that such points are called critical points of f .

Note that f need not have a local extrema at a critical point. The critical points are candidates for local extrema only. In [link] , we show that if a continuous function f has a local extremum, it must occur at a critical point, but a function may not have a local extremum at a critical point. We show that if f has a local extremum at a critical point, then the sign of f switches as x increases through that point.

A function f(x) is graphed. It starts in the second quadrant and increases to x = a, which is too sharp and hence f’(a) is undefined. In this section f’ > 0. Then, f decreases from x = a to x = b (so f’ < 0 here), before increasing at x = b. It is noted that f’(b) = 0. While increasing from x = b to x = c, f’ > 0. The function has an inversion point at c, and it is marked f’(c) = 0. The function increases some more to d (so f’ > 0), which is the global maximum. It is marked that f’(d) = 0. Then the function decreases and it is marked that f’ > 0.
The function f has four critical points: a , b , c , and d . The function f has local maxima at a and d , and a local minimum at b . The function f does not have a local extremum at c . The sign of f changes at all local extrema.

Questions & Answers

find the equation of the tangent to the curve y=2x³-x²+3x+1 at the points x=1 and x=3
Esther Reply
derivative of logarithms function
Iqra Reply
how to solve this question
sidra
ex 2.1 question no 11
khansa
anyone can help me
khansa
question please
Rasul
ex 2.1 question no. 11
khansa
i cant type here
khansa
Find the derivative of g(x)=−3.
Abdullah Reply
any genius online ? I need help!!
Guzorochi Reply
how can i help you?
Pina
need to learn polynomial
Zakariya
i will teach...
nandu
I'm waiting
Zakariya
plz help me in question
Abish
How can I help you?
Tlou
evaluate the following computation (x³-8/x-2)
Murtala Reply
teach me how to solve the first law of calculus.
Uncle Reply
teach me also how to solve the first law of calculus
Bilson
what is differentiation
Ibrahim Reply
only god knows😂
abdulkadir
f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
Naufal Reply
fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
diron
pweding paturo nsa calculus?
jimmy
how to use fundamental theorem to solve exponential
JULIA Reply
find the bounded area of the parabola y^2=4x and y=16x
Omar Reply
what is absolute value means?
Geo Reply
Chicken nuggets
Hugh
🐔
MM
🐔🦃 nuggets
MM
(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
Ismael
find integration of loge x
Game Reply
find the volume of a solid about the y-axis, x=0, x=1, y=0, y=7+x^3
Godwin Reply
how does this work
Brad Reply
Can calculus give the answers as same as other methods give in basic classes while solving the numericals?
Cosmos Reply
log tan (x/4+x/2)
Rohan
please answer
Rohan
y=(x^2 + 3x).(eipix)
Claudia
is this a answer
Ismael
Practice Key Terms 7

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

Ask