Please help me with this i give brainliest.
1 answer:
You might be interested in
Answer:
(a) Decreasing on (0, 1) and increasing on (1, ∞)
(b) Local minimum at (1, 0)
(c) No inflection point; concave up on (0, ∞)
Step-by-step explanation:
ƒ(x) = x² - x – lnx
(a) Intervals in which ƒ(x) is increasing and decreasing.
Step 1. Find the zeros of the first derivative of the function
ƒ'(x) = 2x – 1 - 1/x = 0
2x² - x -1 = 0
( x - 1) (2x + 1) = 0
x = 1 or x = -½
We reject the negative root, because the argument of lnx cannot be negative.
There is one zero at (1, 0). This is your critical point.
Step 2. Apply the first derivative test.
Test all intervals to the left and to the right of the critical value to determine if the derivative is positive or negative.
(1) x = ½
ƒ'(½) = 2(½) - 1 - 1/(½) = 1 - 1 - 2 = -1
ƒ'(x) < 0 so the function is decreasing on (0, 1).
(2) x = 2
ƒ'(0) = 2(2) -1 – 1/2 = 4 - 1 – ½ = ⁵/₂
ƒ'(x) > 0 so the function is increasing on (1, ∞).
(b) Local extremum
ƒ(x) is decreasing when x < 1 and increasing when x >1.
Thus, (1, 0) is a local minimum, and ƒ(x) = 0 when x = 1.
(c) Inflection point
(1) Set the second derivative equal to zero
ƒ''(x) = 2 + 2/x² = 0
x² + 2 = 0
x² = -2
There is no inflection point.
(2). Concavity
Apply the second derivative test on either side of the extremum.
The function is concave up on (0, ∞).
Answer:
Step-by-step explanation:
you can find where the first derivative is 0 to find the critical points
g'(x) = ( x^4 -5x^2 +4)' = 4x³-10x
g'(x) =0, make y =0 to find find where g'(x) is 0
4x³-10x =0 , factor 2x
2x(2x²-5)= 0 , each factor must be 0
2x= 0, so x= 0
2x²-5 =0, so x = ±√5/2
we now have 3 critical points -√5/2, 0, and √5/2
make intervals (-∞, -√5/2), (-√5/2, 0) , (0, √5/2) and (√5/2, +∞)
pick a point to test on each interval: -2, -1, 1 and 2 for example, and
calculate g'(x) = 4x³-10x at those points
for x= -2 we have 4(-2)³-10(-2) = -12 , negative number, decrease
for x= -1 we have 4(-1)³-10(-1) =6, positive number, increase
for x= 1 we have 4(1)³-10(1) = -6, negative number, decrease
for x= 2 we have 4(2)³-10(2) = 12, positive number, increase
we went from a decrease to an increase on intervals (-∞, -√5/2), (-√5/2, 0) so x= - √5/2 ≈ -1.58 is a minimum
we went from a decrease to an increase on intervals (0, √5/2), (√5/2, +∞) so
x= √5/2 ≈ 1.58 is a minimum as well
