Answer to Question #148460 in Calculus for Rishi Francis Roy

Question #148460
Determine whether the following functions
i). f(x) = 2x3 - 3x + 1; [-2; 2]
ii). f(x) = e^x, [0; log 4]
iii). f(x) = log 2x, [1; e]
iv). f(x) = sin^(-1) x, [0; 1/2]
meet the conditions of the Mean Value Theorem on the interval. If so, nd the
point(s) guaranteed to exist by the theorem
1
Expert's answer
2020-12-04T04:57:43-0500

i) "f(x)=2x^3-3x+1" is continuous and differentiable for "x\\in \\R" as polynomial.

Then "f" meets the conditions of the Mean Value Theorem on the interval "[-2, 2]"

"f'(x)=6x^2-3=>f(c)=6c^2-3"

"f(-2)=2(-2)^3-3(-2)+1=-9"

"f(2)=2(2)^3-3(2)+1=11"


"11-(-9)=(6c^2-3)(2-(-2))"

"6c^2-3=5"

"c^2=\\dfrac{4}{3}"

"c_1=-\\dfrac{2\\sqrt{3}}{3}, c_2=\\dfrac{2\\sqrt{3}}{3}"

ii) "f(x)=e^x" is continuous and differentiable for "x\\in \\R" .

Then "f" meets the conditions of the Mean Value Theorem on the interval "[0, \\log4]."

Let "\\log" denotes natural logarithm.

"f'(x)=e^x=>f(c)=e^c"

"f(0)=e^0=1"

"f(\\log4)=e^{\\log4}=4"

"4-1=e^c(\\log4-1)"

"e^c=\\dfrac{3}{\\log4-1}"

"c=\\log(\\dfrac{3}{\\log4-1})"

iii) "f(x)=\\log_2(x)" is continuous and differentiable for "x>0" .

Then "f" meets the conditions of the Mean Value Theorem on the interval "[1, e]."

"f'(x)=\\dfrac{1}{x\\ln2}=>f(c)=\\dfrac{1}{c\\ln2}"

"f(1)=\\log_2(1)=0"

"f(e)=\\log_2(e)=\\dfrac{1}{\\ln2}"

"\\dfrac{1}{\\ln2}-0=\\dfrac{1}{c\\ln2}(e-1)"

"c=e-1"

iv) "f(x)=\\sin^{-1}x" is continuous on "[0, 1\/2]" and differentiable on "(0,1\/2)."

Then "f" meets the conditions of the Mean Value Theorem on the interval "[0, 1\/2]."

"f'(x)=\\dfrac{1}{\\sqrt{1-x^2}}=>f(c)=\\dfrac{1}{\\sqrt{1-c^2}}"

"f(0)=0"

"f(1\/2)=\\dfrac{\\pi}{6}"

"\\dfrac{\\pi}{6}-0=\\dfrac{1}{\\sqrt{1-c^2}}(\\dfrac{1}{2}-0)"

"\\sqrt{1-c^2}=\\dfrac{3}{\\pi}"

"c^2=1-\\dfrac{9}{\\pi^2}, c>0"




"c=\\dfrac{\\sqrt{\\pi^2-9}}{\\pi}"


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