7) Givenf(x) =-1.6x° – 3.r* +10x(a) Plot the function.(b)Use analytical methods to prove that the function is concave for all valuesof x.(c) Differentiate the function and then use a root-location method to solve forthe maximum fx) and the corresponding value of x.

To obtain the solution, use the following conditions. If f''x<0, then the function fx is said to be concave down. If f''x>0, then the function fx is said to be concave up. If f'c=0 and f''c<0, then the function fx attains its maximum value at x=c. If f'c=0 and f''c>0, then the function fx attains its minimum value at x=c.To graph the given function, substitute arbitrary values for x in fx=-1.6x6-3x4+10x and simplify. Tabulate the values of x and fx. Plot the points x,fx in the xy-plane and connect them by a smooth line. fx=-1.6x6-3x4+10xf-1=-1.6-16-3-14+10-1=-14.6f0=-1.606-304+100=0f1=-1.616-314+101=5.4f2=-1.626-324+102=-130.4 x -1 0 1 2 f(x) -14.6 0 5.4 -130.4 Hence, the graph of the function fx=-1.6x6-3x4+10x is as follows.Differentiate the function fx=-1.6x6-3x4+10x with respect to x and simplify, to find f'x and f''x. fx=-1.6x6-3x4+10xf'x=-1.66x5-34x3+101=-9.6x5-12x3+10f''x=-9.65x4-123x2+0=-48x4-36x2 Since f''x<0 for all values of x. Therefore, the function fx=-1.6x6-3x4+10x is concave for all values of x.To determine the maximum value of the function fx=-1.6x6-3x4+10x, solve the equation f'x=0 for x. To obtain the root of the function gx=-9.6x5-12x3+10, apply the Newton-Raphson method. Let the initial guess be x0=0.5. Substitute 0.5 for x in the formula xn+1=xn-gxng'xn, and simplify. Then, continue the process until the x values are repeated. The iteration values are tabulated as follows. gx=-9.6x5-12x3+10gx0=g0.5=-9.60.55-120.53+10=8.2g'x=-48x4-36x2g'x0=g'0.5=-480.54-360.52=-12x1=x0-gx0g'x0=0.5-8.2-12≈1.183333 Iteration x f(x) 0 0.5 8.2 1 1.183333 -32.158381 2 0.960826 -8.505577 3 0.846109 -1.431717 4 0.817687 -0.069754 5 0.816154 -0.000192 6 0.816150 0 Thus, the solution of the equation f'x=0 is x≈0.81615.Substitute x≈0.81615 in the equation fx=-1.6x6-3x4+10x and simplify, to find the maximum value of the given function. f''x=-48x4-36x2f''0.81615=-480.816154-360.816152≈-45.27676fx=-1.6x6-3x4+10xf0.81615=-1.60.816156-30.816154+100.81615≈6.35756 Since f''0.81615<0. So, the function fx=-1.6x6-3x4+10x attains its maximum value at x≈0.81615. Hence, the maximum value of the given function is f0.81615≈6.35756.