- If f:R→R is defined by f (x)=2x+x, then f3x-f-x-4x equals
The correct answer is f:R→R f (x)=2x+x When x≥0,then f (x)=2x+x=3x When x<0, then f (x)=2x-x=x Now, when x≥0 f (3x)-f (-x)-4x=33x--x-4x=9x+x-4x
- If $f:R\\to R$ is defined by $f(x)=x^2-3$, find $f^{-1} (-1)$
If $f:R\to R$ is defined by $f (x)=x^2-3$, find $f^ {-1} (-1)$ My Attempt : $$f (x)=x^2-3$$ $$y=x^2-3$$ Then how to proceed further?
- Let f : R → r be defined by f(x) = 2x + |x|. Then f(2x) + f(-x) – f(x . . .
vote answered Jun 4, 2021 by Zafaa (29 5k points) selected Jun 6, 2021 by rahul01 Best answer Option : (B) f (x) = 2x + |x| f (2x) = 2 (2x)+|2x| = 4x+2|x| f (-x) = 2 (-x)+|-x| f (2x)+f (-x )- f (x) = 4x+2|x|-2x+|-x|- (2x+|x|) = 4x+2|x|-2x+|x|-2x-|x| = 2|x| ← Prev Question Next Question →
- Let f R to R be defined as f (x) left - GeeksforGeeks
Let f : R \to → R be defined as f (x) = \begin {cases} \frac {x^3} { (1 - \cos 2x)^2} \log_e \left ( \frac {1 + 2x e^ {-2x}} { (1 - x e^ {-x})^2} \right), x \neq 0 \\ \alpha, x = 0 \end {cases} f (x) = {(1−cos2x)2x3 loge ((1−xe−x)21+2xe−2x), α, x = 0 x = 0 If f is continuous at x = 0, then I ff iscontinuousatx = 0,then \alpha α
- A function f : R → R defined by f (x) = 2 + x2 is ______. - Mathematics
For one-one, f (x 1) = f (x 2) ⇒ 2 + x 1 2 = 2 + x 2 2 ⇒ x 1 2 = x 2 2 ⇒ x 1 = ±x 2 ⇒ x 1 = x 2 or x 1 = –x 2 Thus, f (x) is not one-one For onto Let f (x) = y such that y ∈ R = x 2 = y – 2 ⇒ x = ± y 2 Put y = –3, we get x = ± 3 2 = ± 5 Which is not possible as root of negative is not a real number Hence, x is not real
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