f(x)=2×3−6x,[0,3]f(x)=2×3-6x,[0,3] Find the absolute extrema of the function on the closed interval.

f(x)=2×3−6x,[0,3]f(x)=2×3-6x,[0,3] Find the absolute extrema of the function on the closed interval.

• MATH

x=e−4x=e-4 Evaluate g(x)=ln(x)g(x)=ln(x) at the indicated value of xx without using a calculator.

Plug in the value of x in g(x)g(x) g(x)=lne−4g(x)=lne-4 Bring down the exponent =−4lne=-4lne Remember that lne=1lne=1 Simplify, g(x)=−4g(x)=-4

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x=e5x=e5 Evaluate g(x)=ln(x)g(x)=ln(x) at the indicated value of xx without using a calculator.

Plug in the value of x for g(x)g(x) g(x)=lne5g(x)=lne5 Bring down the exponent, =5lne=5lne Remember that lne=1lne=1 Simplify, g(x)=5g(x)=5

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e2x=3e2x=3 Write the exponential equation in logarithmic form.

Rewrite into exponential form If Then, loge3=2xloge3=2x or ln3=2xln3=2x

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e−0.9=0.406…e-0.9=0.406… Write the exponential equation in logarithmic form.

Rewrite into exponential form If, Then, loge0.406=−0.9loge0.406=-0.9 or ln0.406=−0.9ln0.406=-0.9

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e12=1.6487…e12=1.6487… Write the exponential equation in logarithmic form.

Rewrite into exponential form If, Then, loge1.6487=(12)loge1.6487=(12) or ln1.6487=(12)ln1.6487=(12)

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e2=7.3890…e2=7.3890… Write the exponential equation in logarithmic form.

Rewrite into exponential form, if e2=7.3890e2=7.3890 Then, loge7.3890=2loge7.3890=2or ln7.3890=2ln7.3890=2

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ln(1)=0ln(1)=0 Write the logarithmic equation in exponential form.

Rewrite using ln rules e0=1e0=1

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ln(250)=5.521…ln(250)=5.521… Write the logarithmic equation in exponential form.

Rewrite using ln rules e5.521=250e5.521=250

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ln(7)=1.945…ln(7)=1.945… Write the logarithmic equation in exponential form.

Rewrite using ln rules e1.945=7e1.945=7

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ln(12)=−0.693…ln(12)=-0.693… Write the logarithmic equation in exponential form.

Rewrite using ln rules e−0.693=12e-0.693=12

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log(5x+3)=log(12)log(5x+3)=log(12) Use the One-to-One Property to solve the equation for xx.

Since you have logs with the same bases on both sides, they cancel leaving you with 5x+3=125x+3=12 Solve for x x=1.8x=1.8

• MATH

log(2x+1)=log(15)log(2x+1)=log(15) Use the One-to-One Property to solve the equation for xx.

Since you have logs with the same bases on both sides, they cancel leaving you with 2x+1=152x+1=15 Solve for x x=7x=7

• MATH

log2(x−3)=log2(9)log2(x-3)=log2(9) Use the One-to-One Property to solve the equation for xx.

Since you have logs with the same bases on both sides, they cancel leaving you with x−3=9x-3=9 Solve for x x=12x=12

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log5(x+1)=log5(6)log5(x+1)=log5(6) Use the One-to-One Property to solve the equation for xx.

Since you have logs with the same bases on both sides, they cancel leaving you with x+1=6x+1=6 Solve for x x=5x=5

• MATH

9log9(15)9log9(15) Use the properties of logarithms to simplify the expression.

There is a very specific rule for this, If x=alogabx=alogab Then, it equals x=bx=bApply to this problem, 9log915=159log915=15

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logπ(π)logπ(π) Use the properties of logarithms to simplify the expression.

The log base of some value to the some value always equals 1. (loga(a)=1loga(a)=1 ) Therefore, this simplifies down to 11

• MATH

log3.2(1)log3.2(1) Use the properties of logarithms to simplify the expression.

The log of any base a of 1 is 0. (loga(1)=0loga(1)=0 ) Therefore, this simplifies down to 00

• MATH

log11(117)log11(117) Use the properties of logarithms to simplify the expression.

Bring down the 7 7log11(11)7log11(11) Note that loga(a)=1loga(a)=1 Therefore, this simplifies down to 77

• MATH

g(x)=logb(x),x=b−3g(x)=logb(x),x=b-3 Evaluate the function at the indicated value of xx without…

Plug in the value of x in g(x)=logb(b−3)g(x)=logb(b-3) Bring down the exponent =−3lobb(b)=-3lobb(b) Note that logb(b)=1logb(b)=1 Therefore, g(x)=−3g(x)=-3

• MATH

g(x)=loga(x),x=a2g(x)=loga(x),x=a2 Evaluate the function at the indicated value of xx without…

Plug in the value of x in g(x)g(x) g(x)=loga(a2)g(x)=loga(a2) =2loga(a)=2loga(a) Note that loga(a)=loga(a)= Therefore, g(x)=2g(x)=2

• MATH

f(x)=log(x),x=10f(x)=log(x),x=10 Evaluate the function at the indicated value of xx without using a…

Plug in the value of x for f(x)f(x) f(x)=log10f(x)=log10 Assume a base of 10 f(x)=log1010f(x)=log1010 Remember that loga(a)=1loga(a)=1 Simplify, f(x)=1f(x)=1

• MATH

f(x)=log8(x),x=1f(x)=log8(x),x=1 Evaluate the function at the indicated value of xx without using…

Plug in the value of x in f(x)f(x) f(x)=log8(1)f(x)=log8(1) Remember that for all loga(1)=0loga(1)=0 simplify, f(x)=0f(x)=0

• MATH

f(x)=log25(x),x=5f(x)=log25(x),x=5 Evaluate the function at the indicated value of xx without using…

Plug in the value of x in f(x)f(x) f(x)=log25(5)f(x)=log25(5) y=log25(5)y=log25(5) 25y=525y=5 52y=5152y=51 2y=12y=1 y=12y=12

• MATH

f(x)=log2(x),x=64f(x)=log2(x),x=64 Evaluate the function at the indicated value of xx without using…

Plug in the value of x in f(x)f(x) f(x)=log2(64)f(x)=log2(64) y=log2(64)y=log2(64) 2y=642y=642y=262y=26 y=6y=6

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240=1240=1 Write the exponential equation in logarithmic form. For example, the logarithmic…

Rewrite in logarithmic form, If, Then, log24(1)=0log24(1)=0

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4−3=1644-3=164 Write the exponential equation in logarithmic form. For example, the…

Rewrite in exponential form If, Then, log4(164)=−3log4(164)=-3

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932=27932=27 Write the exponential equation in logarithmic form. For example, the…

Rewrite in exponential form If, Then, log927=32log927=32

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53=12553=125 Write the exponential equation in logarithmic form. For example, the logarithmic…

Rewrite in exponential form If, Then, log5(125)=3log5(125)=3

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log64(8)=(12)log64(8)=(12) Write the logarithmic equation in exponential form. For example, the…

Rewrite in logarithmic form, If,  log64(8)=(12)log64(8)=(12) Then, 6412=86412=8

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log32(4)=(25)log32(4)=(25) Write the logarithmic equation in exponential form. For example, the…

Rewrite into exponential form If, Then, 3225=43225=4

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log9(181)=−2log9(181)=-2 Write the logarithmic equation in exponential form. For example, the…

Rewrite into exponential form If, Then, 9−2=(181)9-2=(181)

• MATH

log4(16)=2log4(16)=2 Write the logarithmic equation in exponential form. For example, the…

Rewrite into exponential form If log4(16)=2log4(16)=2 Then, 42=1642=16

• MATH

y=tan(πx8),[0,2]y=tan(πx8),[0,2] Find the absolute extrema of the function on the closed interval.

Given the function y=tan(πx8)y=tan(πx8) in [0,2]. We have to find the absolute extrema of the function in the closed interval [0,2]. So first we have to take the derivative of the function and equate…

• MATH

y=3cos(x),[0,2π]y=3cos(x),[0,2π] Find the absolute extrema of the function on the closed interval.

Given: y=3cos(x),[0,2π]y=3cos(x),[0,2π] First find the critical x values of the function. To find the critical values, set the derivative equal to zero and solve for the x value(s). f'(x)=−3sin(x)=0f′(x)=-3sin(x)=0…

• MATH

g(x)=sec(x),[−π6,π3]g(x)=sec(x),[-π6,π3] Find the absolute extrema of the function on the closed interval.

Given the function g(x)=sec(x) in the interval [-pi/6, pi/3] We have to find the absolute extrema of the function on the closed interval. So first let us find the derivative of the function and…

• MATH

f(x)=sin(x),[5π6,11π6]f(x)=sin(x),[5π6,11π6] Find the absolute extrema of the function on the closed…

Given the function: f(x)=sin(x)f(x)=sin(x) in the closed interval [5 pi/6 , 11 pi/6] We have to find the absolute extrema of the function on the closed interval. Step 1: Check whether f(x) is a continuous…

• MATH

h(x)=[[2−x]],[−2,2]h(x)=[[2-x]],[-2,2] Find the absolute extrema of the function on the closed interval.

The function h(x) = [2-x] is the nearest integer function. For values of x lying in the closed set [-2,2], the maximum value of h(x) = 2 – (-2) = 2 + 2 = 4. The minimum value of x is 2 – 2 = 0 The…

• MATH

f(x)=[[x]],[−2,2]f(x)=[[x]],[-2,2] Find the absolute extrema of the function on the closed interval.

f(x) = [x], is the nearest integer function. For values of x in the closed set [-2, 2], the maximum value of f(x) is 2 and the minimum value of f(x) is -2. The absolute extrema of the function are…

• MATH

g(x)=|x+4|,[−7,1]g(x)=|x+4|,[-7,1] Find the absolute extrema of the function on the closed interval.

Hello! For x<=-4 x+4<=0 and g(x) = -(x+4).The minimum on the interval [-7, -4] is at the x=-4, g(-4) = 0.The maximum on the interval [-7, -4] is at the x=-7, g(-7) = 3. For x>=-4…

• MATH

y=3−|t−3|,[−1,5]y=3-|t-3|,[-1,5] Find the absolute extrema of the function on the closed interval.

Hello! Consider this function for t in [-1, 3] and [3, 5] separately. At the first interval t<=3 and |t-3| = -(t-3) = 3-t.So y(t) = 3 – (3-t) = t.The maximum is at the maximum t, i.e. at t=3,…

• MATH

h(t)=tt+3,[−1,6]h(t)=tt+3,[-1,6] Find the absolute extrema of the function on the closed interval.

Given: h(t)=tt+3,[−1,6]h(t)=tt+3,[-1,6] Find the critical values for t by setting the derivative equal to zero and solving for the t value(s). h'(t)=(t+3)(1)−t(1)(t+3)2=0h′(t)=(t+3)(1)-t(1)(t+3)2=0 t+3−t=0t+3-t=0 3=03=0 A critical…

• MATH

h(s)=1s−2,[0,1]h(s)=1s-2,[0,1] Find the absolute extrema of the function on the closed interval.

You need to find the derivative of the function, using the quaotient rule, such that: h'(s)=1’⋅(s−2)−1⋅(s−2)'(s−2)2h′(s)=1′⋅(s-2)-1⋅(s-2)′(s-2)2 h'(s)=0⋅(s−2)−1⋅1(s−2)2h′(s)=0⋅(s-2)-1⋅1(s-2)2 h'(s)=1(s−2)2h′(s)=1(s-2)2 You…

• MATH

f(x)=2xx2+1,[−2,2]f(x)=2xx2+1,[-2,2] Find the absolute extrema of the function on the closed…

You need to evaluate the absolute extrema of the function, hence, you need to differentiate the function with respect to x, using the quotient rule, such that:

• MATH

g(t)=t2t2+3,[−1,1]g(t)=t2t2+3,[-1,1] Find the absolute extrema of the function on the closed…

Given: g(t)=t2t2+3[−1,1]g(t)=t2t2+3[-1,1] Find the critical values for t by setting the derivative equal to zero and solving for the t value(s). g'(t)=(t2+3)(2t)−t2(2t)(t2+3)2=0g′(t)=(t2+3)(2t)-t2(2t)(t2+3)2=0 …

• MATH

g(x)=x−−√3,[−8,8]g(x)=x3,[-8,8] Find the absolute extrema of the function on the closed interval.

You need to find out the absolute extrema of the given function, hence, you need to differentiate the function with respect to x, such that: g'(x)=(x−−√3)’g′(x)=(x3)′

• MATH

y=3×23−2x,[−1,1]y=3×23-2x,[-1,1] Find the absolute extrema of the function on the closed interval.

You need to find out the absolute extrema of the given function, hence, you need to differentiate the function with respect to x, such that: y’=3⋅(23)⋅x23−1−2y′=3⋅(23)⋅x23-1-2 You need to solve for x…

• MATH

f(x)=2×3−6x,[0,3]f(x)=2×3-6x,[0,3] Find the absolute extrema of the function on the closed interval.

You need to find the absolute extrema of the function, hence, you need to differentiate the function with respect to x, such that: f'(x)=(2×3−6x)’f′(x)=(2×3-6x)′ f'(x)=6×2−6f′(x)=6×2-6 You need to solve for x…

• MATH

f(x)=x3−(32)x2,[−1,2]f(x)=x3-(32)x2,[-1,2] Find the absolute extrema of the function on the closed…

Given: f(x)=x3−(32)x2,[−1,2]f(x)=x3-(32)x2,[-1,2] First find the critical x value(s) of the function. To find the critical x value(s), set the derivative equal to zero and solve for the x value(s)….

• MATH

h(x)=5−x2,[−3,1]h(x)=5-x2,[-3,1] Find the absolute extrema of the function on the closed interval.

Given: h(x)=5−x2,[−3,1]h(x)=5-x2,[-3,1] Find the critical values of the function by setting the derivative equal to zero and solving for x. When the derivative is equal to zero the slope of the tangent line…