Applications of Derivatives

The 1st deÑetes1 Applications of Derivatives We will use the derivatives to find the following in this chapter: • The intervals where a function is increasing or decreasing using the 1st derivative test. • The extreme values or the relative maximum or minimum values of a function, using the 1st and the 2nd derivative tests. • Solve Optimization problems, where you must either maximize or minimize a function subject to certain constraints. • learn how to recover a function from its derivative. Find an antiderivative of a function. (An introduction to the Integral Calculus) Applications of the first derivative ✗ i f- , ✗ y < ✗2 f { ✗ , a z y, Y > Iz , decreasing function critical points The function 𝑓 is increasing over (−∞, −2) and (2, ∞). 𝑓 is a decreasing function over (−2, 2). < increasing function an , , An Algebraic Method to determine Intervals Where a Given Function is Increasing or Decreasing: f- T if f ' > 0 . f- Is if f- 1<0 f- - Guidelines for finding intervals on which a function is increasing or decreasing: • Find the critical values of the function by determining the 𝑥 − values where either 𝑓 ′ (𝑥 ) = 0 𝑜𝑟 𝑓 ′ (𝑥 ) is not defined. • Use the critical values to divide the 𝑥 −axis into intervals and choose a test value in each interval. • Find the sign of 𝑓 ′ (𝑥 ) for each test value chosen in step 2 and use this information to determine where 𝑓(𝑥) is increasing or decreasing and to classify any extrema as relative maxima or minima. • If 𝑓 ′ (𝑐) > 0 𝑡ℎ𝑒𝑛 𝑓(𝑥 ) is an increasing function on that interval, and If 𝑓 ′ (𝑐) < 0 𝑡ℎ𝑒𝑛 𝑓(𝑥 ) is a decreasing function on that interval. Examples: Determine the intervals where the function is increasing and where it is decreasing. 1. 𝑓 (𝑥 ) = 𝑥 3 − 12𝑥 − 5 ' f- ( X) = 0 3×2-12=0 31×2-47=0 3( ✗ 2) ( ✗ +27=0 - critical numbers The ✗ = 2 , ✗ = -2 off : ← (-9-2) - sign off 'm ' f- CX )=3( ✗ 3 > o -271×-14 • (2,0) (-2/2) 0 3 co 70 :¥Ilu ]1:D HII"J|:÷I I ' g- < o f- Torch C- 44 f ' > 0 2. 𝑓(𝑥 ) = 𝑥 3 − 3𝑥 2 − 24𝑥 + 32 : ✗ - C- a. a) values where I > f- ' f4✗7=0 fk✗7failstoexis 1×7=31,2-6¥ -213=0-3 31×2--2×-81=0 ✗ =Y , Ñ+f=o - ears test value 3l→} fix) 131×-411×+22 - - E. -3 t < 0 = -2 * 5 34141 > 0 ltEE.IT/Eeu:J1ftoren-M# Y¥YEi f 3 31-4714 ✗ ' > 0 0th c- a-↳ 3. 𝑓 (𝑥 ) = 2 𝑥 4 − 2𝑥 3 − 6𝑥 2 + 8 1 ✗ 4. 𝑓(𝑥 ) = 𝑥 + 𝑥 cd f' ' 1=0 values off 1×1=1-1×2 : =µ-?÷ ×¥z = critical values affix) : f- 4×7=0 ✗ - values where flex)DNE of 111=4-1,02 - (-1/0) -1<(-0-5) (-9-1) test values, ' sign 9- f- Ix ) -2 ( ( NJ il ) ) ¥6.5 2- to 70 - C3)( . =¥¥ > f ' > o 0 over (-4-1) < / lftorer-a.rs# 4¥Y 0 E;u¥÷] The 1st Derivative Test: |f'lÑ ✗ ¢ , fcc ) ) ✗ =c ¢ x=c , fees ) /fk✗ - 1. Use the 1st derivative test to determine the intervals where the function is increasing and where it is decreasing. Find any relative extrema. 𝑓(𝑥 ) = 4𝑥 3 − 9𝑥 2 − 30𝑥 + 25 -51×7=4×3 9×2-30×+25 f- 0,0) rititf |f4x - ( : ¥ " 2×2-3×-5=0 9--2 c= -5 -3 , b= , I = ✗ |×=E - -b±÷ / =3±H; ✗ 31¥ = ¥=a • = ) -5 0 3 . . -2 value . <0 f4✗7 70 =6( 2.x? - r , Ks ÷fau€ test , -1 )¥, 3.x -5 . 20 k % flower ¥%IY¥÷IT¥ l←• V / f- T ohh 25 - By the f N 1st 1 has a f= f has at a ( = ( 1 f- 1 max , 42) min f- (2-5) ) 2- 5, . ft )) , relative 2.5 , test , relative at value 2. g- -1 derivative ° -43-75) . value 3 2. Find the open intervals on which 𝑓 (𝑥 ) = 𝑥 3 − 𝑥 2 is increasing or 2 decreasing. Find the local extrema of the function using the 1st derivative test. Solution: Note that 𝑓 is differentiable on the entire real number line. To determine the critical numbers of 𝑓, set 𝑓 ′ (𝑥 ) equal to zero. 3 3 2 2 𝑓(𝑥 ) = 𝑥 3 − 𝑥 2 => 𝑓 ′ (𝑥 ) = 3𝑥 2 − (2𝑥 ) = 0 => 3(𝑥 )(𝑥 − 1) = 0 => 𝑥 = 0, 1 Interval Test the three intervals determined by these two critical numbers. −∞ < 𝑥 < 0 0<𝑥<1 𝑥 = −1 Test Value Sign of 𝑓 ′ (𝑥) Conclusion 𝑓 ′ (−1) = 6 > 0 Increasing 1 2 1 3 𝑓′ ( ) = − < 0 2 4 Decreasing 𝑥= 1<𝑥<∞ 𝑥=2 𝑓 ′ (2) = 6 > 0 Increasing So, 𝑓 is increasing on the interval (−∞, 0) and(1, ∞) and decreasing on (0, 1). Relative Extrema/ The First Derivative Test 3 The function 𝑓(𝑥 ) = 𝑥 3 − 2 𝑥 2 has a relative maximum at the point (0, 0) because 𝑓 is increasing immediately to the left of 𝑥 = 0 and decreasing immediately to the right of 𝑥 = 0. Similarly, 𝑓 has a relative minimum at the 1 point (1, − 2) because 𝑓 is decreasing immediately to the left of 𝑥 = 1 and increasing immediately to the right of 𝑥 = 1.