Consider the quadratic function g(x)=1x2-4x+36 Define a new function S(x) by S(x

Question

Consider the quadratic function g(x)=1x2-4x+36 Define a new function S(x) by S(x)=the s lope of the diagonal line through g(x)- . (a) Find the positive critical value of S(x) x= 12 (b) Determine whether S(x) is concave up or concave down at the critical value you found in (a) and use the Second Derivative Test to determine if the critical value gives a local maximum or a local minimum. S(x) is concave upat the critical value. Therefore, the critical value gives a local minimumof Sx) (c) If x must be positive, what is the lowest possible value of S(x)? (d) If x is between 1 and 6, what are the largest and smallest possible values of S(x)? largest value of S(x): 32.25 smallest value of S(x) : 3.5 (e) Give the longest interval over which S(x) is decreasing and g(x) is increasing from x= 8 to x= 12

Explanation / Answer

Answer is S(12)=(1÷4)(12) - 4 + (36÷12)

=3 - 4 + 3

=2

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