Please Answer this question correctly. Example question below.

Let f be the function given by f (x) = 5x 2. What is the approximation for f (24.95) found by using the line tangent to the graph of f at a = 25? Tangent Line: y = Approximation: f (24.95) ~ When v the second derivative v , meaning the graph of f is near the point of tangency, making the value ??? an Submit AnswerLet f be the function given by f (x) - 5x 2. What is the approximation for f (24.95) found by using the line tangent to the graph of f at x = 25? Find point of tangency: 5 f(25) = 5(25) = 1 V25 Point of tangency: (25, 1) Find the derivative: f (ac) = 5x 5 -5 f' (a) = --2 or 2 Use the power rule Find derivative at x = 25: The slope of the tangent line -5 -5 5 f (25) = 2V25 2(125) 250 50 Plug in and simplify Slope of tangent line: 50 Write equation of tangent line 1 y- 1 = 50 ( - 25) y = 50 (x - 25) + 1 Solve for y1 = 25 1 31 50(33 )+ Approximate f(21.95): 1 = 24.9525 1 y 50( )+ 1 = D.U5 1 y 50( )+ "y = 0.001 + 1 = 1.001 The linear approximation for f[24.95] is 1.001. To see whether the found value is an under approximation or an over approximation, we must analyze the second derivative. f'(.r,) = %$ NAIF-H 15 . 1 firm) 2 Ia: E or 5 41? When .1: f} l], the second derivative will be positive, meaning the graph of f is concave up near the point of tangenov, making the value 1.001 an under approximation. 1 Tangent Line: 1; = %($ 25) | 1 Solve for y Solve for y Solve for 3; From before Just a reminder Let f be the function given by f($] = 235% . 1;What is the approximation for f(3.95] found by using the line tangent to the graph of f at m = 4? Approximation: f[3.95) E |:| "When - the second derivative - meaning the graph of f is near the point of tangenev, making the value ??? an :- Submit lewe r