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If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1?

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Bunuel
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If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   10 Jun 2020, 06:56
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If \(f(x) = A(x−B)^3\), where \(AB≠0\), what is value of x for which \(f(x) = 1\)?


A. \(B - \sqrt[3]{A}\)

B. \(B - \frac{1}{\sqrt[3]{A}}\)

C. \(B + \sqrt[3]{A}\)

D. \(-B + \frac{1}{\sqrt[3]{A}}\)

E. \(B + \frac{1}{\sqrt[3]{A}}\)
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E
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Re: If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   10 Jun 2020, 07:10
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F(x) = A(x-B)^3
For which value f(x) will be equal to 1
That means
A(x-B)^3 = 1
(x-B)^3 = 1/A
Taking cube root on both sides.
(X-B) = 1/³√A
X = B+1/³√A

Verification: put X = B+1/³√A
F(x) = A(x-B)^3
F(x) = A(B+1/³√A-B)^3
F(x) = A(1/A)
F(x) = 1

Answer is E

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Re: If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   10 Jun 2020, 07:18
Given = F(x) = A(x-B)^3
F(x) = 1
x=?

since F(x)=1, we can substitute this value in the equation F(x) = A(x-B)^3
therefore A(x-B)^3=1
(x-B)^3 = 1/A
(x-B) = ∛(1/ A)
x-B = 1/∛A
x = (1/∛A) + B

hence the correct choice is option E.
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If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   10 Jun 2020, 07:54
We can also take the approach as follows:

1. For f(x) = 1, we need the expression in the bracket to be equal to 1/A.
2. (x-B)^3, looking at the expression we need +B to negate B and rest in as the cube root of 1/A.
3. Hence the answer is B+1/(A)^1/3 (Option E)

This solution might be a different approach but saves :)
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Re: If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   10 Jun 2020, 22:14
f(x) = A (x - B)^3 ; A*B not equal to "0"
x= ? when f(x) = 1

f(x) = A(x- B)^3 = 1
=> (x-B)^3 = 1/ A
=> x-B = 1 / (A)^1/3
=> x = 1/ (A)^1/3 + B


Answer : E
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Re: If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   10 Jun 2020, 22:32
If f(x)=A(x−B)^3, where AB≠0, what is value of x for which f(x)=1?


A. B−A√3B−A3

B. B−1A√3B−1A3

C. B+A√3B+A3

D. −B+1A√3−B+1A3

E. B+1A√3

SOL: given: a, b != 0 and f(x)=1

f(x) = A(x-B)^3 = 1
=> (x-B)^3 = 1/A (taking cube root on both sides)
=> (x-B) = (1/A)^1/3
=> x = B + (1/A)^1/3

Ans is E
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Re: If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   12 Jun 2020, 14:38
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Bunuel wrote:
If \(f(x) = A(x−B)^3\), where \(AB≠0\), what is value of x for which \(f(x) = 1\)?


A. \(B - \sqrt[3]{A}\)

B. \(B - \frac{1}{\sqrt[3]{A}}\)

C. \(B + \sqrt[3]{A}\)

D. \(-B + \frac{1}{\sqrt[3]{A}}\)

E. \(B + \frac{1}{\sqrt[3]{A}}\)


Solution:

A(x - B)^3 = 1

(x - B)^3 = 1/A

x - B = 3^√(1/A)

x = B + 3^√(1/A) = B + 1/(3^√A)

Answer: E
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If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   12 Jun 2020, 19:13
Bunuel wrote:
If \(f(x) = A(x−B)^3\), where \(AB≠0\), what is value of x for which \(f(x) = 1\)?


A. \(B - \sqrt[3]{A}\)

B. \(B - \frac{1}{\sqrt[3]{A}}\)

C. \(B + \sqrt[3]{A}\)

D. \(-B + \frac{1}{\sqrt[3]{A}}\)

E. \(B + \frac{1}{\sqrt[3]{A}}\)


Asked: If \(f(x) = A(x−B)^3\), where \(AB≠0\), what is value of x for which \(f(x) = 1\)?

\(f(x) = A(x-B)^3 = 1\)
\((x-B)^3 = 1/A\)
\(x = B + \frac{1}{\sqrt[3]{A}\)

IMO E
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If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink] New post   06 Jul 2020, 12:35
\(f(x) = A(x−B)^3=1\)
\((x−B)^3=\frac{1}{A}\)
\(x−B=\frac{1}{\sqrt[3]{A}}\)
\(x=B-\frac{1}{\sqrt[3]{A}}\)

Answer E
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If f(x) = A(x−B)^3, where AB≠0, what is value of x for which f(x) = 1? [#permalink]
06 Jul 2020, 12:35
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