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求解 2x+5/6-x+2/3leqx-2/9+1/2(x-3) Microsoft Math Solver

Web$$ f:\mathbb R^{+}\to [-5,\infty) $$ $$ f(x)=9x^{2}+6x-5 $$ $$ f^{-1}(y)=({\frac{(\sqrt{y+6})-1}{3}}) $$ Now I have to show that $ f $ is invertible with $f^{-1}$ WebProve that the function f: N → N, defined by f(x) =x2+x+1 is one-one but not onto. Find inverse of f: N → S, where S is range of f. Solution Given f(x) =x2+x+1 To prove one-one, we will show if f(x1) =f(x2), then x1 = x2 Let f(x1) =f(x2) ⇒ x2 1+x1+1= x2 2+x2+1 ⇒ x2 1+x1−x2 2−x2= 0 ⇒ (x1−x2)(x1+x2+1) =0 Since x1+x2+1 ≠0 ∀ x1,x2 ∈N, we have x1−x2 =0 recipes using chicken mince australia

Consider f : R+ → [−5, ∞), given by f(x) = 9x2 + 6x − 5. Show that f …

WebApr 3, 2024 · Best answer To Show: that f is invertible To Find: Inverse of f [NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)] one-one function: A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different images in B. Web1 Now I have to show that is invertible with I'm trying to show that and I have done to some extend: Now from here it is clear that it it won't be equal to . Kindly tell me where I'm getting wrong , is my process wrong or there is any calculation mistake. functions Share Cite Follow edited Mar 7, 2014 at 5:38 asked Mar 7, 2014 at 4:43 Singh WebOct 3, 2024 · Here, function f : R + →[−5,∞) is given as f(x) = 9x 2 + 6x − 5.. Let y be any arbitrary element of [−5,∞). Let y = 9x 2 + 6x − 5 . ⇒ y = (3x + 1) 2 − 1 − 5 = (3x + 1) 2 − 6 ⇒ (3x + 1) 2 = y + 6 ⇒ (3x + 1) 2 = √(y + 6) [as y ≥ − 5 ⇒ y + 6 ≥ 0] x = (√(y = 6) - 1)/3. Therefore, f is onto, thereby range f = [−5,∞). unshare a shared workbook