Consider the function below. (Round the answers to two decimal places. If you ne

Question

Consider the function below. (Round the answers to two decimal places. If you need to use - or , enter -INFINITY or INFINITY.)

f(x) = 7x tan(x)
-/2 < x < /2

(a) Find the interval where the function is increasing.
(  ,  )

Find the interval where the function is decreasing.
(  ,  )

(b) Find the local minimum value.


(c) Find the interval where the function is concave up.
(  ,  )

(d) Use this information to sketch the graph of f. (Do this on paper. Your instructor may ask you to turn in this graph.)

Explanation / Answer

f(x) = 7xtanx

f'(x) = 7tanx + 7xsec2x =0 => 7(tanx + xsec2x)=0 => 7(tanx+x(1+tan2x)) =0 => tanx+x+xtan2x =0 =>

xtan2x + tanx + x=0 => x(1+tan2x) = -tanx => x=0

hence the intervals are (-pi/2,0) U (0,pi/2)

f"(x) = 7sec2x + 7sec2x + 14xsecxsecxtanx = 14sec2x + 14xsec2xtanx =0 => 14sec2x (1+tanx)=0

=> sec2x=0 or tanx=-1 => x=-pi/4

Consider (-pi/2,pi/4)

let x= -pi/8, f"(-pi/8) = 14(1.000023)2(-0.0068) <0

Now consider `(pi/4,pi/2)

Let x=pi/8, f"(pi/8) = 14(0.0000023)2(0.0068) >0

Consider (-pi/2,0)

Let x= -pi/4 , f'(-pi/4) = 7(-1)-7pi/4(1/2) = -7-7pi/8 <0

Consider (0,pi/2)

Let x=pi/4, f'(pi/4) = 7(1)+7(pi/4)(1/2) = 7+7pi/8 >0

(a) the function f(x) is increasing on (0,pi/2) because derivative of f is positive in this interval

The function is decreasing on (-pi/2,0) because derivative of f is negative in this interval

(b) The local minimum value is given by f(0) = 0

(c) Hence the function is concave up on (pi/4,pi/2) because f"(x) is positive in that interval

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