the function ( f(x)=x^{3}+14 x^{2}+56 x + 74 ) is graphed below. plot a line segment connecting the points…

the function ( f(x)=x^{3}+14 x^{2}+56 x + 74 ) is graphed below. plot a line segment connecting the points on ( f ) where ( x=-9 ) and ( x=-4 ). afterwards, determine all values of ( c ) which satisfy the conclusion of the mean value theorem for ( f ) on the closed interval ( -9 leq x leq -4 ). plot a line by clicking in two locations. click the line to delete it.

the function ( f(x)=x^{3}+14 x^{2}+56 x + 74 ) is graphed below. plot a line segment connecting the points on ( f ) where ( x=-9 ) and ( x=-4 ). afterwards, determine all values of ( c ) which satisfy the conclusion of the mean value theorem for ( f ) on the closed interval ( -9 leq x leq -4 ). plot a line by clicking in two locations. click the line to delete it.

Answer

Explanation:

Step1: Calculate ( f(-9) ) and ( f(-4) )

[ \begin{align*} f(-9)&=(-9)^{3}+14(-9)^{2}+56(-9)+74\ &=-729 + 14\times81-504 + 74\ &=-729+1134-504 + 74\ &=-25 \end{align*} ] [ \begin{align*} f(-4)&=(-4)^{3}+14(-4)^{2}+56(-4)+74\ &=-64+14\times16-224 + 74\ &=-64 + 224-224+74\ &=10 \end{align*} ]

Step2: Find the slope of the secant line

The slope ( m=\frac{f(-4)-f(-9)}{-4-(-9)}=\frac{10 - (-25)}{-4 + 9}=\frac{35}{5}=7 )

Step3: Find the derivative of ( f(x) )

( f^{\prime}(x)=3x^{2}+28x + 56 )

Step4: Set ( f^{\prime}(c)=7 )

[ \begin{align*} 3c^{2}+28c + 56&=7\ 3c^{2}+28c+49&=0 \end{align*} ] Using the quadratic formula ( c=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} ) with ( a = 3 ), ( b=28 ), ( c = 49 ) [ \begin{align*} c&=\frac{-28\pm\sqrt{28^{2}-4\times3\times49}}{2\times3}\ &=\frac{-28\pm\sqrt{784 - 588}}{6}\ &=\frac{-28\pm\sqrt{196}}{6}\ &=\frac{-28\pm14}{6} \end{align*} ] [ c_{1}=\frac{-28 + 14}{6}=\frac{-14}{6}=-\frac{7}{3}, \quad c_{2}=\frac{-28-14}{6}=\frac{-42}{6}=-7 ] Since ( -9\leq c\leq -4 ), we check the values. ( c=-7 ) is in the interval ( [-9,-4] ) and ( c =-\frac{7}{3}\approx - 2.33) is not in the interval ( [-9,-4] )

Answer:

( c=-7 )