WebAnswer (1 of 7): y=5x^2–4x+3 dy/dx=10x-4 For maximum or minimum putting dy/dx=0 10x-4=0=> x= 2/5 d^2y/dx^2= 10 (+ve) There exist minimum at x=2/5 Minimum value = … WebA function basically relates an input to an output, there’s an input, a relationship and an output. For every input...
Solve Quadratic equations 5x^2-2x-1=0 Tiger Algebra Solver
Weby= 5x^2 – 2x + 1. Differntiate it once and equate to zero. So, dy/dx = 10x -2. For minimum value. 10x-2=0. Or, x= 1/5. So, y= 5x1/25 – 2x1/5 + 1. = 1/5-2/5+ 1. = 4/5. WebJun 23, 2016 · Minimum value: − 13 4 Explanation: A parabola (with a positive coefficient for x2) has a minimum value at the point where its tangent slope is zero. That is when XXXdy dx = d(x2 +5x +3) dx = 2x +5 = 0 which implies XXXx = − 5 2 Substituting − 5 2 for x in y = x2 +5x + 3 gives XXXy = ( − 5 2)2 + 5( − 5 2) +3 XXXy = 25 4 − 25 2 + 3 summary of the folk of the faraway tree
The minimum value if the polynomial p(x) = 3x^2 - 5x + 2 is - Toppr
WebMay 17, 2024 · The given equation is Quadratic, which means it only has 1 maximum or 1 minimum point. 1. Re-arranging the equation, f (x) = −2x2 +x − 1 Since the coefficient of the term x2 is a negative, the graph of the curve has a maximum point. Conversely, if it is a positive, it has a minimum point. graph {-2x^2+x-1 [-4.896, 5.104, -4.66, 0.34]} WebTo find critical points of a function, take the derivative, set it equal to zero and solve for x, then substitute the value back into the original function to get y. Check the second derivative test to know the concavity of the function at that point. What is a critical point in a function? Webllustration 10. The minimum value of y = 5x2 - 2x + 1 is Solution Verified by Toppr Video Explanation Solve any question of Application of Derivatives with:- Patterns of problems > … pakistan surrounding countries