If x is real the minimum value of x2-8x+17 is
Webx2-8x+17=0 Two solutions were found : x = (8-√-4)/2=4-i= 4.0000-1.0000i x = (8+√-4)/2=4+i= 4.0000+1.0000i Step by step solution : Step 1 :Trying to factor by splitting the middle term ... x2-9x-16x=0 Two solutions were found : x = 25 x = 0 Step by step solution : Step 1 : Step 2 :Pulling out like terms : 2.1 Pull out like factors : x2 - 25x ... Web20 dec. 2024 · Show that the maximum value of (1/x)x is e1/e. LIVE Course for free. Rated by 1 million+ students Get app now Login. Remember. Register; ... Let M and m be respectively the absolute maximum and the absolute minimum values of the function, asked Jan 19, 2024 in Limit, ... (17.6k) Artificial Intelligence (AI) (1.4k) Information ...
If x is real the minimum value of x2-8x+17 is
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Webx2-8x+17=0 Two solutions were found : x = (8-√-4)/2=4-i= 4.0000-1.0000i x = (8+√-4)/2=4+i= 4.0000+1.0000i Step by step solution : Step 1 :Trying to factor by splitting the … Web29 mrt. 2024 · If x is real, the minimum value of x 2 – 8x + 17 is (A)–1 (B) 0 (C) 1 (D) 2 This question is similar to Ex 6.5, 1 (i) - Chapter 6 Class 12 - Application of Derivatives
WebThe minimum value of any quadratic expression ax 2 + bx + c = 0, (when a > 0) is -D/4a Where, D = b 2 - 4ac Calculation: Here, a, b and c is 4, -5 and 1 respectively So, D = (-5) … WebThe maximum value of −x 2−8x+17 is A 17 ` B −1 C 33 D 2 Medium Solution Verified by Toppr Correct option is C) f(x)=−x 2−8x+17 f(x)=−2x−8=0 x=−4 f(x)=−2(−ve) ∴ at x=−4→ …
Web23 apr. 2024 · Solution : To find the minimum points we find the first derivative and get the critical points then substitute that points in the second derivative. Let Derivate w.r.t x, For … WebWhen x is positive, the minimum value of x x is A e e B e 1/e C e −1/e D (1/e) Medium Solution Verified by Toppr Correct option is C) Solve any question of Application of Derivatives with:- Patterns of problems > Was this answer helpful? 0 0 Similar questions If x>0 then the minimum value of x x is Medium View solution >
Webf (x) = x 2 − 8 x + 17 ⇒ f ′ (x) = 2 x − 8 For minimum or maximum we have f ′ (x) = 0 ⇒ 0 = 2 x − 8 ⇒ x = 4 Now f ′′ (x) = 2 At x = 4, f ′′ (x) is + v e ∴ x = 4 is local minima point. ∴ …
WebIf x is real, the minimum value of x 2 – 8x + 17 is 1. Explanation: Let f (x) = x 2 – 8x + 17 f' (x) = 2x – 8 For local maxima and local minima, f' (x) = 0 ∴ 2x – 8 = 0 ⇒ x = 4 So, x = 4 … rises in teachers pensions contributions ukWebLet y=x 2−8x+17=(x 2−2⋅4⋅x+4 2)+1=(x−4) 2+1We know that if x is a real number then x 2≥0⇒(x−4) 2≥0Hence minimum value of y is 0+1=1. rise single epic music 下载WebIf x is real, the minimum value of x2 – 8x + 17 is (A) –1 (B) 0 (C) 1 (D) 2 application of derivative class-12 Share It On Facebook Twitter Email 1 Answer +2 votes answered … rise sign and printWeb23 dec. 2024 · If P = (1, 1), Q = (3, 2) and R is a point on x-axis, then the value of PR + RQ will be minimum at asked Dec 23, 2024 in Limit, continuity and differentiability by Vikky01 ( 42.0k points) applications of derivatives rises in spanishWeb9 aug. 2024 · If x is real, the minimum value of x^2 – 8x + 17 is A. –1 B. 0 C. 1 D. 2. asked Sep 19, 2024 in Derivatives by Chandan01 (51.5k points) applications of derivatives; class-12; Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. rise singleissue parent the week articlesWebIf x is real, find the maximum and minimum values of x 2+2x+3x 2+14x+9. Medium Solution Verified by Toppr Let y= x 2+2x+3x 2+14x+9. ⇒y(x 2+2x+3)=x 2+14x+9 ⇒(y−1)x 2+2(y−7)x+3y−9=0 Since, x is real, so discriminant of above equation will be greater than or equal to 0. D≥0 ⇒4(y−7) 2−4(y−1)(3y−9)≥0 ⇒(y−7) 2−(y−1)(3y−9)≥0 ⇒y 2+49−14y−3(y … rise skin creamWebLet f(x) = x2 − 8x + 17 Differentiating both sides with respect to x, we get f ' x = 2 x-8 For maxima or minima, f ' x = 0 ⇒ 2 x-8 = 0 ⇒ x = 4 Now, f ' ' x = 2 > 0 So, x = 4 is the point … rise show cast