Assume f' is given by the graph in the figure. Suppose f is continuous and that f(0) = 0.(a) Find f(3) =(b) Find f(7)=(c) Find[²849-8f'(x) dx12WebAssign Plot345f(x)67X

The antiderivative of a function f( x) is a function F(x), such as d/dx (F(x)) = f( x). The general antiderivative fiction is F(x) + C because d/dx (F (x) +C) = f(x), C is constantTo solve a definite integral, we use the Fundamental theorem of Calculus part 2∫a b f(x) dx = F(x) |a b = F(b) - F( a), where F(x) is an antiderivative of f(x) : d/dx (F(x) ) = f(x). Any differentiable function is continuous, but not any continuous function is differentiable. To check if a function f(x) is continuous at a point a, we verify the following relationshiplimx ->a- f(x) = limx->a+ f(x) = f(a) Knowing that the function f(x) is continuous and f( 0) = 0 and the graph of it's derivative is shown in the plot, we can accurately find f(x) by integrating the derivative function. The derivative function can be written step-wise asf'(x) = 4 0 < x < 2 -4 2 < x < 3 8 3 < x < 4 -8 4 < x < 6 4 6 < x < 7f(x) = 4x + C1 0 < x < 2 -4x + C2 2 < x < 3 8x + C3 3 < x < 4 -8x + C4 4 < x < 6 4x + C5 6 < x < 7For the function be continuous we impose the continuity property and use f(0) =0f(0) = 0limx->0- f(x) = limx->0+ = 0limx->0+ 4x + C1 = 0 C1 = 0limx->2- f(x) = limx->2+ f(x) => limx->2- (4x + C1) = limx->2+ (-4x + C2 ) => 8 + C1 = - 8 + C2 => 8 + 0 = - 8 + C2 C2 = 16limx->3- f(x) = limx->3+ f(x) => limx->3- (-4x + C2 ) = limx->3+ (8x + C3 ) => -12 + 16 = 24 + C3 C3 = -20limx->4- (8x + C3 ) = limx->4+ (-8x + C4 ) => 32 + (-20) = - 32 + C 4 C4 = 44limx->6- (-8x + C4) = limx->6+ (4x + C5 ) => - 48 + 44 = 24 + C5 C5 = -28f(x) = 4x 0 ≤ x < 2 -4x + 16 2 ≤ x < 3 8x - 20 3 ≤ x < 4 -8x + 44 4 ≤ x < 6 4x - 28 6 ≤ x ≤ 7a) f(3) = 8 × 3 - 20 = 24 - 3 = 21b) f(7) = 4 × 7 - 28 = 28 - 28 = 0c) ∫0 7 = f(7) - f(0) = 0 - 0 = 0a) 21b) 0c) 0