• MATH

g(x)=x13−x−23g(x)=x13-x-23 Find the critical numbers of the function

Given: g(x)=x13−x−23g(x)=x13-x-23 Find the critical value(s) of the function by setting the first derivative equal to zero and solving for the x value(s). g'(x)=(13)x−23+(23)x−53=0g′(x)=(13)x-23+(23)x-53=0…

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• MATH

h(t)=t34−2t14h(t)=t34-2t14 Find the critical numbers of the function

Given: h(t)=t34−2t14h(t)=t34-2t14 Find the critical number(s) by setting the first derivative equal to zero and solving for the x value(s). h'(t)=(34)t−14−(14)(2)t−34=0h′(t)=(34)t-14-(14)(2)t-34=0…

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• MATH

h(p)=p−1p2+4h(p)=p-1p2+4 Find the critical numbers of the function

Given: h(p)=p−1p2+4h(p)=p-1p2+4 Find the critical value(s) by setting the first derivative equal to zero and solving for the x value(s). Find the derivative by using the quotient rule….

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• MATH

g(y)=y−1y2−y+1g(y)=y-1y2-y+1 Find the critical numbers of the function

You need to find the critical points of the function, hence, you need to evaluate the solutions to the equation g'(y) = 0. You need to evaluate the first derivative, using the quotient rule:…

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• MATH

g(t)=|3t−4|g(t)=|3t-4| Find the critical numbers of the function

The modulus of the equation of the function is defined as it follows: |3t−4|=3t−4if3t−4≥0⇒t≥43|3t-4|=3t-4if3t-4≥0⇒t≥43 or |3t−4|=4−3tift<43|3t-4|=4-3tift<43 You need to evaluate the critical…

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• MATH

g(t)=t4+t3+t2+1g(t)=t4+t3+t2+1 Find the critical numbers of the function

g(t)=t4+t3+t2+1g(t)=t4+t3+t2+1 or,g'(t)=4t3+3t2+2tor,g′(t)=4t3+3t2+2t or,g'(t)=t(4t2+3t+2)or,g′(t)=t(4t2+3t+2) g'(t)=0g′(t)=0 (4t2+3t+2)t=0(4t2+3t+2)t=0Hence the critical number is t = 0

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• MATH

f(x)=2×3+x2+2xf(x)=2×3+x2+2x Find the critical numbers of the function

You need to evaluate the critical numbers of the function, hence, you need to evaluate the soutions to the first derivative, such that: f'(x)=0f′(x)=0 f'(x)=(2×3+x2+2x)f′(x)=(2×3+x2+2x)

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• MATH

f(x)=2×3−3×2−36xf(x)=2×3-3×2-36x Find the critical numbers of the function

You need to evaluate the critical numbers of the function, hence, you need to evaluate the soutions to the first derivative, such that: f'(x)=0f′(x)=0 f'(x)=(2×3−3×2−36x)f′(x)=(2×3-3×2-36x)

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• MATH

f(x)=x3+6×2−15xf(x)=x3+6×2-15x Find the critical numbers of the function

You need to evaluate the critical numbers of the function, hence, you need to evaluate the soutions to the first derivative, such that: f'(x)=0f′(x)=0 f'(x)=(x3+6×2−15x)f′(x)=(x3+6×2-15x)

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• MATH

f(x)=4+(13)x−(12)x2f(x)=4+(13)x-(12)x2 Find the critical numbers of the function

f(x)=4+(13)x−(12)x2f(x)=4+(13)x-(12)x2 or,f(x)=24+2x−3x2or,f(x)=24+2x-3×2 or,f'(x)=2−6xor,f′(x)=2-6x Thus, for critical numbers f'(x) = 0 or, 2-6x = 0 or, x = 1/3 is the critical number

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• MATH

y=sec2(x)x2+1y=sec2(x)x2+1 Find the limit, if possible

limx→∞sec2(xx2+1)limx→∞sec2(xx2+1) limx→∞sec2(x)x2(1+1×2)limx→∞sec2(x)x2(1+1×2) limx→∞sec2⎛⎝⎜1x(1+1×2)⎞⎠⎟limx→∞sec2(1x(1+1×2)) plug in the value of x to evaluate the limit =sec2(1∞(1+0))=sec2(1∞(1+0)) =sec2(0)==sec2(0)=…

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• MATH

y=3x−sin2(x)y=3x-sin2(x) Find the limit, if possible

limx→03x−sin2xlimx→03x-sin2x plug in the value of x =3⋅0−sin2(0)=3⋅0-sin2(0) =0 limx→∞3x−sin2xlimx→∞3x-sin2x limx→∞x(3−sin2xx)limx→∞x(3-sin2xx) Apply the squeeze theorem to evaluate the limit of sin^2x/x…

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• MATH

y=x1−x2−−−−−√y=x1-x2 Find the limit, if possible

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→∞x(1−x2−−−−−√)=(∞)(1−∞−−−−−√)limx→∞x(1-x2)=(∞)(1-∞) Since the result is indeterminate, you need to multiply and…

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• MATH

y=9−x2−−−−−√y=9-x2 Find the limit, if possible

limx→09−x2−−−−−√limx→09-x2 plug in the value of x to evaluate the limit = 9−02−−−−−√=39-02=3 limx→∞9−x2−−−−−√limx→∞9-x2 =9−∞2−−−−−−√9-∞2 limit does not exist

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• MATH

y=x−−√+1x−−√y=x+1x Find the limit, if possible

This function is defined for x>0. When x tends to +0, sqrt(x) tends to 0 and 1/sqrt(x) tends to +infinity, and y tends to +infinity. The same answer for x tending to infinity. And for x tending…

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• MATH

y=x+12x−1y=x+12x-1 Find the limit, if possible

limx→∞x+12x−1limx→∞x+12x-1 or,limx→∞(xx)+(1x)(2xx)−(1x)or,limx→∞(xx)+(1x)(2xx)-(1x) or,limx→∞1+(1x)2−(1x)=12or,limx→∞1+(1x)2-(1x)=12

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• MATH

y=csc(2x)y=csc(2x) Find the limit, if possible

If you evaluate the limit of function from the left, as x→0−x→0- yields: limx→0−csc(2x)=limx→0−1sin(2x)limx→0-csc(2x)=limx→0-1sin(2x) Replacing 2sinx⋅cosx2sinx⋅cosx for sin(2x)sin(2x)yields:

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• MATH

y=3x23y=3×23 Find the limit, if possible

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→∞3×23=3⋅∞2−−−√3=∞limx→∞3×23=3⋅∞23=∞ Hence, evaluating the given limit, for x→∞x→∞ , yields…

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• MATH

y=3×2−4y=3×2-4 Find the differential dy of the given function.

For a function y=f(x) the differential dy is f'(x)dx. Here dy=6x⋅dxdy=6x⋅dx .

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• MATH

limx→∞x−cos(x)xlimx→∞x-cos(x)x Find the limit.

x−cos(x)x=1−cos(x)x.x-cos(x)x=1-cos(x)x. When, cos(x) is bounded (by 1) and 1/x -> 0. Therefore cos(x)/x -> 0 and the answer is 1.

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• MATH

limx→∞sin(2x)xlimx→∞sin(2x)x Find the limit.

The numerator, sin(2x), is bounded (by 1), while the denominator, x, tends to infinity, and 1/x tends to zero. Therefore the limit in question is 0.

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• MATH

limx→∞cos(1x)limx→∞cos(1x) Find the limit.

When x→∞x→∞ , 1/x -> 0 and cos(1/x) -> cos(0)=1 (cosx is continuous at 0).

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• MATH

limx→∞12x+sin(x)limx→∞12x+sin(x) Find the limit.

Divide numerator and denominator by x and obtain 1×2+sin(x)x.1×2+sin(x)x. As x tends to infinity, 1/x tends to zero and sin(x) is bounded. The limit is 0/(2+0)=0.

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• MATH

limx→−∞2x(x6−1)13limx→-∞2x(x6-1)13 Find the limit.

You need to evaluate the limit, hence, you need to replace -∞∞ for x in expression under the limit: limx≻∞2xx6−1−−−−−√3=−∞∞limx≻∞2xx6-13=-∞∞ Since the limit is indeterminate, you need…

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• MATH

limx→∞x+1(x2+1)13limx→∞x+1(x2+1)13 Find the limit.

limx→∞x+1×2+1−−−−−√3limx→∞x+1×2+13 limx→∞x(1+1x)x2(1+1×2)−−−−−−−−−√3limx→∞x(1+1x)x2(1+1×2)3 limx→∞x(1+1x)x231+1×2−−−−−−√3limx→∞x(1+1x)x231+1×23 limx→∞x13(1+1x)1+1×2−−−−−−√3⎞⎠⎟limx→∞x13(1+1x)1+1×23) apply…

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• MATH

limx→−∞x4−1−−−−−√x3−1limx→-∞x4-1×3-1 Find the limit.

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→−∞x4−1−−−−−√x3−1=∞4−1−−−−−−√(−∞)3−1=∞−∞limx→-∞x4-1×3-1=∞4-1(-∞)3-1=∞-∞ Since the result is…

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• MATH

limx→∞x2−1−−−−−√2x−1limx→∞x2-12x-1 Find the limit.

limx→∞x2−1−−−−−√2x−1limx→∞x2-12x-1 or,limx→∞(x2x2)−(1×2)−−−−−−−−−−−√(2xx)−(1x)or,limx→∞(x2x2)-(1×2)(2xx)-(1x) or,limx→∞1−0−−−−−√2−0=1–√2=12or,limx→∞1-02-0=12=12

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• MATH

limx→∞5×2+2×2+3−−−−−√limx→∞5×2+2×2+3 Find the limit.

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→∞5×2+2×2+3−−−−−√=5∞+2∞+3−−−−−√=∞∞limx→∞5×2+2×2+3=5∞+2∞+3=∞∞ Since the result is…

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• MATH

limx→−∞2x+1×2−x−−−−−−√limx→-∞2x+1×2-x Find the limit.

Transform the equation under lim, 2x+1×2−x−−−−−−√2x+1×2-x . Divide the numerator and denominator by |x|, take into account that x2=|x|2×2=|x|2: 2x+1×2−x−−−−−−√=2⋅x|x|+1|x|1−x|x|2−−−−−−√2x+1×2-x=2⋅x|x|+1|x|1-x|x|2…

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• MATH

limx→−∞xx2+x−−−−−−√limx→-∞xx2+x Find the limit.

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→−∞xx2+x−−−−−−√=−∞∞+(−∞)−−−−−−−−−−√limx→-∞xx2+x=-∞∞+(-∞) Since the result is indeterminate, you need to factor…

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• MATH

limx→−∞xx2−x−−−−−−√limx→-∞xx2-x Find the limit.

Let’s divide the numerator and denominator by |x|. For negative x x/|x|=-1 and x^2/|x|^2=1. So we obtain (-1)/sqrt(1-x/|x|^2). Now we can let x tend to minus infinity, x/|x|->0 and the entire…

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• MATH

limx→−∞x3−4×2+1limx→-∞x3-4×2+1 Find the limit.

Divide both numerator and denominator by x2x2 : x3−4×2+1=x−4×21+1×2.×3-4×2+1=x-4×21+1×2.When x→−∞,x→-∞, 1×2→0.1×2→0. Therefore the limit is equal to −∞−01+0=−∞.-∞-01+0=-∞.

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• MATH

limx→−∞5x2x+3limx→-∞5x2x+3 Find the limit.

Since this is a −∞−∞-∞-∞ form Thus, applying L’Hospital’s rule we get limx→−∞5x2x+3=limx→−∞(10x)=−∞limx→-∞5x2x+3=limx→-∞(10x)=-∞

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• MATH

limx→∞5×3+110×3−3×2+7limx→∞5×3+110×3-3×2+7 Find the limit.

limx→∞5×3+110×3−3×2+7limx→∞5×3+110×3-3×2+7 or,limx→∞(5x3x3)+(1×3)(10x3x3)−(3x2x3)+(7×3)or,limx→∞(5x3x3)+(1×3)(10x3x3)-(3x2x3)+(7×3)

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• MATH

limx→∞xx2+3limx→∞xx2+3 Find the limit.

Divide both numerator and denominator by x: xx2+3=1x+3xxx2+3=1x+3x . When x→∞,x→∞, 3x→03x→0 and the limit is 1∞+0=1∞1∞+0=1∞ = 0.

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• MATH

limx→−∞4×2+5×2+3limx→-∞4×2+5×2+3 Find the limit.

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→−∞4×2+5×2+3=4∞+5∞+3=∞∞limx→-∞4×2+5×2+3=4∞+5∞+3=∞∞ Since the result is indeterminate, you need…

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• MATH

limx→∞2x−13x+2limx→∞2x-13x+2 Find the limit.

limx→∞2x−13x+2limx→∞2x-13x+2 or,limx→∞(2xx)−(1x)(3xx)+(2x)or,limx→∞(2xx)-(1x)(3xx)+(2x) or,limx→∞2−(1x)3+(2x)=2−03+0=23or,limx→∞2-(1x)3+(2x)=2-03+0=23

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• MATH

limx→−∞(5x−x3)limx→-∞(5x-x3) Find the limit.

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→−∞(5x−x3)=5−∞−−∞3=0+∞limx→-∞(5x-x3)=5-∞–∞3=0+∞ Since the result is indeterminate, you need bring the…

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• MATH

limx→∞(4+3x)limx→∞(4+3x) Find the limit.

limx→∞4+(3x)=4+0=4limx→∞4+(3x)=4+0=4

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• MATH

limx→∞5x324x−−√+1limx→∞5x324x+1 Find the limit, if possible

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→∞5x324x12+1=5∞4∞+1=∞∞limx→∞5x324x12+1=5∞4∞+1=∞∞ Since the result is indeterminate, you need…

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• MATH

limx→∞5x324x32+1limx→∞5x324x32+1 Find the limit, if possible

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→∞5x324x32+1=5∞4∞+1=∞∞limx→∞5x324x32+1=5∞4∞+1=∞∞ Since the result is indeterminate, you…

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• MATH

limx→∞5x324x2+1limx→∞5x324x2+1 Find the limit, if possible

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→∞5x324x2+1=5∞4∞+1=∞∞limx→∞5x324x2+1=5∞4∞+1=∞∞ Since the result is indeterminate, you need to…

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• MATH

limx→∞5−2x323x−4limx→∞5-2x323x-4 Find the limit, if possible

You need to evaluate the limit, hence, you need to replace oo for x in equation: limx→∞5−2x323x−4=5−∞3∞−4=−∞∞limx→∞5-2x323x-4=5-∞3∞-4=-∞∞ Since the result is indeterminate, you need…

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• MATH

limx→∞5−2x323x32−4limx→∞5-2x323x32-4 Find the limit, if possible

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→∞5−2x323x32−4=5−∞3∞−4=−∞∞limx→∞5-2x323x32-4=5-∞3∞-4=-∞∞ Since the result is indeterminate,…

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• MATH

limx→∞5−2x323x2−4limx→∞5-2x323x2-4 Find the limit, if possible

You need to evaluate the limit, hence, you need to replace ∞∞ for x in equation: limx→∞5−2x323x2−4=5−2⋅∞3∞−4=−∞∞limx→∞5-2x323x2-4=5-2⋅∞3∞-4=-∞∞ Since the result is indeterminate,…

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• MATH

limx→∞3−2x23x−1limx→∞3-2x23x-1 Find the limit, if possible

Using L’ Hospital’s Rule limx→∞3−(2×2)3x−1limx→∞3-(2×2)3x-1 or,limx→∞−4x3or,limx→∞-4×3 or,limx→∞−4×3=−∞or,limx→∞-4×3=-∞

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• MATH

limx→∞3−2x3x−1limx→∞3-2x3x-1 Find the limit, if possible

limx→∞3−2x3x−1limx→∞3-2x3x-1 or,limx→∞(3x)−(2xx)(3xx)−(1x)or,limx→∞(3x)-(2xx)(3xx)-(1x) or,limx→∞0−23−0=−23or,limx→∞0-23-0=-23

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• MATH

limx→∞3−2x3x3−1limx→∞3-2x3x3-1 Find the limit, if possible

limx→∞3−2x3x3−1limx→∞3-2x3x3-1 or,limx→∞(3×3)−(2xx3)(3x3x3)−(1×3)or,limx→∞(3×3)-(2xx3)(3x3x3)-(1×3) or,limx→∞(3×3)−(2×2)3−(1×3)=0−03−0=0or,limx→∞(3×3)-(2×2)3-(1×3)=0-03-0=0

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• MATH

limx→∞x2+2x−1limx→∞x2+2x-1 Find the limit, if possible

limx→∞x2+2x−1limx→∞x2+2x-1 or,limx→∞2x1or,limx→∞2×1 or,limx→(2x)=∞or,limx→(2x)=∞

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• MATH

limx→∞x2+2×2−1limx→∞x2+2×2-1 Find the limit, if possible

limx→∞x2+2×2−1limx→∞x2+2×2-1 or,limx→∞(x2x2)+(2×2)(x2x2)−(1×2)or,limx→∞(x2x2)+(2×2)(x2x2)-(1×2) or,limx→∞1+(2×2)1−(1×2)=1+01−0=1or,limx→∞1+(2×2)1-(1×2)=1+01-0=1

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