f(x)=sin(x)−1,0<x<2πf(x)=sin(x)-1,0<x<2π
- MATH
f(x)=(12)x4+2x3f(x)=(12)x4+2×3 Find the points of inflection and discuss the concavity of the…
Given: f(x)=(12)x4+2x3f(x)=(12)x4+2×3 Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s). f'(x)=2×3+6x2f′(x)=2×3+6×2 f”(x)=6×2+12x=0f′′(x)=6×2+12x=0…
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- MATH
f(x)=−x3+6×2−5f(x)=-x3+6×2-5 Find the points of inflection and discuss the concavity of the…
Given: f(x)=−x3+6×2−5f(x)=-x3+6×2-5 Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s). f'(x)=−3×2+12xf′(x)=-3×2+12x f”(x)=−6x+12=0f′′(x)=-6x+12=0…
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- MATH
f(x)=x3−6×2+12xf(x)=x3-6×2+12x Find the points of inflection and discuss the concavity of the…
Given: f(x)=x3−6×2+12xf(x)=x3-6×2+12x Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s). f'(x)=3×2−12x+12f′(x)=3×2-12x+12 f”(x)=6x−12=0f′′(x)=6x-12=0…
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- MATH
y=x+2sin(x)(−π,π)y=x+2sin(x)(-π,π) Determine the open intervals on whcih the graph is concave…
This function is undefined at 0, i.e. we have to consider it on (−π,0)(-π,0)and (0,π)(0,π) separately. To determine concavity, we have to find the second derivative and its sign.
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- MATH
y=2x−tan(x),(−π2,π2)y=2x-tan(x),(-π2,π2) Determine the open intervals on whcih the graph is…
y=2x−tanxy=2x-tanx differentiating, y’=2−sec2(x)y′=2-sec2(x) differentiating again, y”=−2sec(x)sec(x)tan(x)y′′=-2sec(x)sec(x)tan(x) y”=−2sec2(x)tan(x)y′′=-2sec2(x)tan(x) y”=−2(1cos2(x))(sin(x)cos(x))y′′=-2(1cos2(x))(sin(x)cos(x)) y”=−2sin(x)cos2(x)y′′=-2sin(x)cos2(x) In…
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- MATH
h(x)=x2−12x−1h(x)=x2-12x-1 Determine the open intervals on which the graph is concave upward or…
h(x)=x2−12x−1h(x)=x2-12x-1 differentiating by applying the quotient rule, h'(x)=(2x−1)(2x)−(x2−1)(2)(2x−1)2h′(x)=(2x-1)(2x)-(x2-1)(2)(2x-1)2 h'(x)=4×2−2x−2×2+2(2x−1)2h′(x)=4×2-2x-2×2+2(2x-1)2 h'(x)=2×2−2x+2(2x−1)2h′(x)=2×2-2x+2(2x-1)2differentiating again,…
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- MATH
g(x)=x2+44−x2g(x)=x2+44-x2 Determine the open intervals on whcih the graph is concave…
g(x)=x2+44−x2g(x)=x2+44-x2 differentiating by applying quotient rule, g'(x)=(4−x2)(2x)−(x2+4)(−2x)(4−x2)2g′(x)=(4-x2)(2x)-(x2+4)(-2x)(4-x2)2 g'(x)=8x−2×3+2×3+8x(4−x2)2g′(x)=8x-2×3+2×3+8x(4-x2)2 g'(x)=16x(4−x2)2g′(x)=16x(4-x2)2differentiating again…
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- MATH
y=−3×5+40×3+135x270y=-3×5+40×3+135×270 Determine the open intervals on whcih the graph is…
y=−3×5+40×3+135x270y=-3×5+40×3+135×270 differentiating y’=1270(−15×4+120×2+135)y′=1270(-15×4+120×2+135) differentiating again, y”=1270(−60×3+240x)y′′=1270(-60×3+240x) y”=−60270(x3−4x)y′′=-60270(x3-4x) y”=−60270x(x+2)(x−2)y′′=-60270x(x+2)(x-2) In order to…
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- MATH
f(x)=x2+1×2−1f(x)=x2+1×2-1 Determine the open intervals on which the graph is concave upward or…
f(x)=x2+1×2−1f(x)=x2+1×2-1 differentiating by applying quotient rule, f'(x)=(x2−1)(2x)−(x2+1)(2x)(x2−1)2f′(x)=(x2-1)(2x)-(x2+1)(2x)(x2-1)2 f'(x)=2×3−2x−2×3−2x(x2−1)2f′(x)=2×3-2x-2×3-2x(x2-1)2 f'(x)=−4x(x2−1)2f′(x)=-4x(x2-1)2differentiating again,…
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- MATH
f(x)=2x23x2+1f(x)=2x23x2+1 Determine the open intervals on whcih the graph is concave…
f(x)=2x23x2+1f(x)=2x23x2+1 differentiating by applying quotient rule, f'(x)=(3×2+1)(4x)−(2×2)(6x)(3×2+1)2f′(x)=(3×2+1)(4x)-(2×2)(6x)(3×2+1)2 f'(x)=12×3+4x−12×3(3×2+1)2f′(x)=12×3+4x-12×3(3×2+1)2 f'(x)=4x(3×2+1)2f′(x)=4x(3×2+1)2 differentiating…
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- MATH
f(x)=24×2+12f(x)=24×2+12 Determine the open intervals on whcih the graph is concave upward or…
Hello! To determine where a function is concave upward or downward, we may use the second derivative. This function is defined everywhere, differentiable and second differentiable everywhere, so…
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- MATH
Given: f(x)=x5−5x+2f(x)=x5-5x+2 Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s). f'(x)=5×4−5f′(x)=5×4-5 f”(x)=20×3=0f′′(x)=20×3=0 x3=0x3=0 x=0x=0…
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- MATH
Given: f(x)=−x3+6×2−9x−1f(x)=-x3+6×2-9x-1 Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s). f'(x)=−3×2+12x−9f′(x)=-3×2+12x-9 f”(x)=−6x+12=0f′′(x)=-6x+12=0…
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- MATH
h(x)=12x−x3h(x)=12x-x3 Determine the open intervals on whcih the graph is concave upward or…
Given: h(x)=12x−x3h(x)=12x-x3 Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s). h'(x)=12−3x2h′(x)=12-3×2 h”(x)=−6x=0h′′(x)=-6x=0 x=0x=0 The…
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- MATH
g(x)=x2−2x−8g(x)=x2-2x-8 Determine the open intervals on whcih the graph is concave upward or…
Given: g(x)=x2−2x−8g(x)=x2-2x-8 Find the critical values for x by setting the second derivative of the function equal to zero and solving for the x value(s). g'(x)=2x−2g′(x)=2x-2 g”(x)=2g′′(x)=2 The critical value for…
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- MATH
f(x)=sin(x)1+cos2(x)f(x)=sin(x)1+cos2(x) Consider the function on the interval (0, 2pi). Find the…
You need to determine the relative extrema of the function, hence, you need to find the solutions to the equation f'(x) = 0. You need to evaluate the derivative of the function, using the quotient…
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- MATH
f(x)=sin2(x)+sin(x)f(x)=sin2(x)+sin(x) Consider the function on the interval (0, 2pi). Apply first…
You need to evaluate the relative extrema of the function, hence, you need to find the solutions to the equation f'(x)=0f′(x)=0 . You need to determine the first derivative, using the chain rule, such…
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- MATH
You need to find the relative extrema of the function, hence, you need to remember that the roots of the equation f'(x) = 0 are the extrema of the function. You need to find the first derivative of…
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- MATH
f(x)=cos2(2x)f(x)=cos2(2x) Consider the function on the interval (0, 2pi). Find the open intervals…
You need to find the open intervals on which the function is increasing or decreasing, hence, you need to find where the derivative is positive or negative, so, you need to evaluate the first…
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- MATH
f(x)=x+2sin(x)f(x)=x+2sin(x) Consider the function on the interval (0, 2pi). Find the open…
This function is differentiable on (0,2π).(0,2π). It is increasing where its derivative is positive and is decreasing when the derivative is negative. f'(x)=1+2cos(x),f′(x)=1+2cos(x), its roots on (0,2π)(0,2π)…
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- MATH
f(x)=sin(x)+cos(x)f(x)=sin(x)+cos(x) Consider the function on the interval (0, 2pi). Find the open…
Given: f(x)=sin(x)+cos(x),(0,2π)f(x)=sin(x)+cos(x),(0,2π) Find the critical values by setting the first derivative equal to zero and solving for the x value(s). f'(x)=cos(x)−sin(x)=0f′(x)=cos(x)-sin(x)=0 cos(x)−sin(x)=0cos(x)-sin(x)=0…
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- MATH
This function is continuous and differentiable on the given interval. It is increasing when f'(x)>0 and decreasing when f'(x)<0. Let’s find f'(x). f(x) = sin(x)*cos(x) + 5 = (1/2)*sin(2x) +…
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- MATH
f(x)=x2+cos(x)f(x)=x2+cos(x) Consider the function on the interval (0, 2pi). Find the open…
This functions is infinitely differentiable. To find its extrema we have to solve the equation f'(x)=0: f'(x)=1/2 – sin(x). This is zero at x=(-1)^k*pi/6+k*pi. At the given interval (0, 2pi) there…
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- MATH
f(x)=x2−2x+1x+1f(x)=x2-2x+1x+1 Find the critical numbers, open intervals on which the…
Given: f(x)=x2−2x+1x+1f(x)=x2-2x+1x+1 Find the critical numbers by setting the first derivative equal to zero and solving for the x value(s). f'(x)=(x+1)(2x−2)−(x2−2x+1)(1)(x+1)2=0f′(x)=(x+1)(2x-2)-(x2-2x+1)(1)(x+1)2=0…
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- MATH
f(x)=x2x2−9f(x)=x2x2-9 Find the critical numbers, open intervals on which the function is…
Given: f(x)=x2x2−9f(x)=x2x2-9 Find the critical value(s) of the function by setting the first derivative equal to zero and solving for the x value(s). f'(x)=(x2−9)(2x)−(x2)(2x)(x2−9)2=0f′(x)=(x2-9)(2x)-(x2)(2x)(x2-9)2=0…
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- MATH
f(x)=xx−5f(x)=xx-5 Find the critical numbers, open intervals on which the function is…
Given: f(x)=xx−5f(x)=xx-5 Find the critical numbers by setting the first derivative equal to zero and solving for the x value(x) f'(x)=(x−5)(1)−x(1)(x−5)2=0f′(x)=(x-5)(1)-x(1)(x-5)2=0 x−5−x=0x-5-x=0 −5=0-5=0 No solution. A…
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- MATH
f(x)=2x+1xf(x)=2x+1x Find the critical numbers, open intervals on which the function is…
Given: f(x)=2x+(1x)=2x+1x−1f(x)=2x+(1x)=2x+1x-1 Find the critical values of x by setting the derivative of the function equal to zero and solving for the x-value(s). f'(x)=2−1×2=0f′(x)=2-1×2=0 2=1×22=1×2 2×2=12×2=1…
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- MATH
f(x)=|x+3|−1f(x)=|x+3|-1 Find the critical numbers, open intervals on which the function is…
Given the function f(x)=|x+3|-1: Rewrite as a piecewise function: f(x)=⎧⎩⎨⎪⎪⎡⎣⎢x+2−1−x−4x>−3x=−3x<−3⎤⎦⎥f(x)={[x+2x>-3-1x=-3-x-4x<-3] The function is continuous on the reals. The first derivative fails to…
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- MATH
f(x)=5−|x−5|f(x)=5-|x-5| Find the critical numbers, open intervals on which the function is…
This function is continuous everywhere and is differentiable everywhere except x=5. For x<=5, f(x)=x, f'(x)=1 and for x>-5 f(x)=10-x, f'(x)=-1. So there are no points for which f'(x)=0, and…
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- MATH
f(x)=(x−3)13f(x)=(x-3)13 Find the critical numbers, open intervals on which the function is…
You need to evaluate the critical values, hence, in order to do so, you must solve for x the equation f'(x)=0f′(x)=0 . You need to find f'(x),f′(x),such that:
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- MATH
f(x)=(x+2)23f(x)=(x+2)23 Find the critical numbers, open intervals on which the function is…
You need to find the open intervals on which the function is increasing or decreasing, hence, you need to find where the derivative is positive or negative, so, you need to evaluate the first…
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- MATH
f(x)=x23−4f(x)=x23-4 Find the critical numbers, open intervals on which the function is…
Given: f(x)=x23−4f(x)=x23-4 Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s). f'(x)=(23)x−13=0f′(x)=(23)x-13=0 23×13=023×13=0 2=02=0 Because 2=0 is not…
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- MATH
f(x)=x13+1f(x)=x13+1 Find the critical numbers, open intervals on which the function is…
Given: f(x)=x13+1f(x)=x13+1 Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s). f'(x)=(13)x−23=0f′(x)=(13)x-23=0 13×23=013×23=0 1=01=0 1=0 is not a…
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- MATH
f(x)=x4−32x+4f(x)=x4-32x+4 Find the critical numbers, open intervals on which the function is…
Given: f(x)=x4−32x+4f(x)=x4-32x+4 Find the critical numbers by setting the first derivative equal to zero and solving for the x value(s). f'(x)=4×3−32=0f′(x)=4×3-32=0 4×3=324×3=32 x3=8×3=8 x=2x=2 The critical value is…
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- MATH
f(x)=x5−5x5f(x)=x5-5×5 Find the critical numbers, open intervals on which the function is…
Given: f(x)=x5−5×5=(15)x5−xf(x)=x5-5×5=(15)x5-x Find the critical x values by setting the derivative equal to zero and solving for the x value(s). f'(x)=x4−1=0f′(x)=x4-1=0 (x2+1)(x2−1)=0(x2+1)(x2-1)=0 x=1,x=−1x=1,x=-1 The critical…
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- MATH
f(x)=(x+2)2(x−1)f(x)=(x+2)2(x-1) Find the critical numbers, open intervals on which the function is…
This function is defined everywhere, its critical numbers are where f'(x)=0. f'(x)=2*(x+2)*(x-1) + (x+2)^2 = (x+2)*(2x-2+x+2) = (x+2)*(3x). This is =0 at x=-2 and x=0. Also f'(x) is negative for x…
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- MATH
f(x)=(x−1)2(x+3)f(x)=(x-1)2(x+3) Find the critical numbers, open intervals on which the function is…
Given: f(x)=(x−1)2(x+3)f(x)=(x-1)2(x+3) Find the critical values by setting the first derivative equal to zero and solving for the x value(s). f'(x)=(x−1)2(1)+(x+3)(2(x−1))=0f′(x)=(x-1)2(1)+(x+3)(2(x-1))=0 x2−2x+1+(x+3)(2x−2)=0x2-2x+1+(x+3)(2x-2)=0…
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- MATH
f(x)=x3−6×2+15f(x)=x3-6×2+15 Find the critical numbers, open intervals on which the function is…
Given: f(x)=x3−6×2+15f(x)=x3-6×2+15 Find the critical values for x by setting the first derivative of the function equal to zero and solving for the x value(s). f'(x)=3×2−12x=0f′(x)=3×2-12x=0 3x(x−4)=03x(x-4)=0 x=0,x=4x=0,x=4…
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- MATH
f(x)=−2×3+3×2−12xf(x)=-2×3+3×2-12x Find the critical numbers, open intervals on which the function…
Critical points are where derivation is equal to zero i.e. f'(x)=0f′(x)=0Therefore, we will calculate function derivative. f'(x)=−6×2+6x−12f′(x)=-6×2+6x-12 Hence we need to find solution to the following…
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- MATH
f(x)=−3×2−4x−2f(x)=-3×2-4x-2 Find the critical numbers, open intervals on which the function is…
Given: f(x)=−3×2−4x−2f(x)=-3×2-4x-2 Find the critical values for x by setting the first derivative of the function equal to zero and solving for the x value(s). f'(x)=−6x−4=0f′(x)=-6x-4=0 −6x=4-6x=4 x=4−6x=4-6 x=−23x=-23…
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- MATH
f(x)=−2×2+4x+3f(x)=-2×2+4x+3 Find the critical numbers, open intervals on which the function is…
Given: f'(x)=−2×2+4x+3f′(x)=-2×2+4x+3 Find the critical values for x by setting the first derivative of the function equal to zero and solving for the x value(s). f'(x)=−4x+4=0f′(x)=-4x+4=0 −4x=−4-4x=-4 x=1x=1 The…
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- MATH
f(x)=x2+6x+10f(x)=x2+6x+10 Find the critical numbers, open intervals on which the function is…
Given: f(x)=x2+6x+10f(x)=x2+6x+10 Find the critical values for x by setting the first derivative of the function equal to zero and solving for the x value(s). f'(x)=2x+6=0f′(x)=2x+6=0 f'(x)=2x=−6f′(x)=2x=-6 x=−3x=-3 The…
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- MATH
f(x)=x2−4xf(x)=x2-4x Find the critical numbers, open intervals on which the function is…
Given: f(x)=x2−4xf(x)=x2-4x Find the critical values of x by setting the derivative of the function equal to zero and solving for the x-value(s). f'(x)=2x−4=0f′(x)=2x-4=0 2x=42x=4 x=2x=2 The critical value is at…
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- MATH
f(x)=sin2(x)+sin(x),0<x<2πf(x)=sin2(x)+sin(x),0<x<2π Identify the open intervals on which the…
This function is differentiable and it is increasing when its derivative is positive (and decreasing when derivative is negative). f'(x)=2sin(x)cos(x) + cos(x)=cos(x)*(1+2sin(x)). This is zero…
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- MATH
y=x−2cos(x),0<x<2πy=x-2cos(x),0<x<2π Identify the open intervals on which the function is…
You need to determine the intervals on which the function increases or decreases, hence, you need to determine where the first derivative is positive or negative. You need to evaluate the first…
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- MATH
h(x)=cos(x2),0<x<2πh(x)=cos(x2),0<x<2π Identify the open intervals on which the function is…
You need to determine the intervals on which the function is increasing or decreasing, hence, you need to find out intervals on which the first derivative is positive or negative. You need to…
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- MATH
f(x)=sin(x)−1,0<x<2πf(x)=sin(x)-1,0<x<2π Identify the open intervals on which the function…
Given f(x)=sin(x)-1: A function is increasing on an interval if it is differentiable on the interval and the first derivative is positive; decreasing if the derivative is negative. f'(x)=cos(x)…
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- MATH
y=x+9xy=x+9x Identify the open intervals on which the function is increasing or decreasing.
This function is defined everywhere except x=0 and is differentiable. To determine where it is increasing or decreasing, compute the derivative: y'(x)=1−9×2.y′(x)=1-9×2. It is positive in
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- MATH
y=x16−x2−−−−−−√y=x16-x2 Identify the open intervals on which the function is increasing or…
We need to take the derivative of the function first in order to identify the critical points or critical numbers, which we can use for the endpoints of our intervals. For the right side we need to…
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- MATH
h(x)=12x−x3h(x)=12x-x3 Identify the open intervals on which the function is increasing or…
You need to find the open intervals on which the function is increasing or decreasing, hence, you need to find where the derivative is positive or negative, so, you need to evaluate the first…
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