Q77 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
Let $f(x)=a+(x-4)^{\dfrac{4}{9}}$, then minima of $f(x)$ is
- Correct answerAa
- Ba $-$ 4
- CNone of these
- D4
Solution
The function is given by $f(x) = a + (x - 4)^{\dfrac{4}{9}}$.
The term $(x - 4)^{\dfrac{4}{9}}$ can be written as $\sqrt[9]{(x - 4)^4}$. Since the exponent $\dfrac{4}{9}$ has an even numerator, the expression $(x - 4)^4$ is always non-negative for all real $x$.
The minimum value of $(x - 4)^4$ is $0$, which occurs at $x = 4$.
Consequently, the minimum value of $(x - 4)^{\dfrac{4}{9}}$ is $0^{\dfrac{4}{9}} = 0$.
Therefore, the minimum value of $f(x) = a + (x - 4)^{\dfrac{4}{9}}$ is $a + 0 = a$.
Answer: a
The term $(x - 4)^{\dfrac{4}{9}}$ can be written as $\sqrt[9]{(x - 4)^4}$. Since the exponent $\dfrac{4}{9}$ has an even numerator, the expression $(x - 4)^4$ is always non-negative for all real $x$.
The minimum value of $(x - 4)^4$ is $0$, which occurs at $x = 4$.
Consequently, the minimum value of $(x - 4)^{\dfrac{4}{9}}$ is $0^{\dfrac{4}{9}} = 0$.
Therefore, the minimum value of $f(x) = a + (x - 4)^{\dfrac{4}{9}}$ is $a + 0 = a$.
Answer: a
Hence, the correct answer is option: 'A'.