Question #7ca7c

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Question #7ca7c

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Answer 1 · 2017-01-08T19:13:53

$8, - 6, 0, undefined$ $8,-6,0," undefined"$

Explanation:

$(a) (f+g)(1) means evaluate f(x)+g(x) when x = 1$ $"(a) (f+g)(1) means evaluate f(x)+g(x) when x = 1"$

$That is (f (1) + g (1)$ $"That is " (f(1)+g(1)$

From the above table for x = 1 , check along the row and read the values given for f(x) and g(x)

$x = 1 \to f (x) = 7 and g (x) = 1$ $x=1tocolor(red)(f(x)=7)" and " color(blue)(g(x)=1)$

$\Rightarrow (f + g) (1) = 7 + 1 = 8$ $rArr(f+g)(1)=7+1=8$

$(b) Similarly$ $" (b) Similarly"$

$(f \times g) (- 1) This time we want the product of the functions when x = - 1$ $(fxxg)(-1)" This time we want the product of the functions when x = - 1"$

From the table for x = - 1 , check along the row and read the values.

$x = - 1 \to f (x) = 3 and g (x) = - 2$ $x=-1tocolor(red)(f(x)=3)" and " color(blue)(g(x)=-2)$

$\Rightarrow (f \times g) (- 1) = 3 \times - 2 = - 6$ $rArr(fxxg)(-1)=3xx-2=-6$

$(c) again similar to above$ $" (c) again similar to above"$

$x = 3 \to f (x) = 9 and g (x) = 9$ $x=3tocolor(red)(f(x)=9)" and " color(blue)(g(x)=9)$

$\Rightarrow (f - g) (3) = 9 - 9 = 0$ $rArr(f-g)(3)=9-9=0$

$(d) x = 0 \to f (x) = 5 and g (x) = 0$ $" (d) " x=0tocolor(red)(f(x)=5)" and " color(blue)(g(x)=0$

$\Rightarrow (\frac{f}{g}) (0) = \frac{5}{0} which is undefined$ $rArr(f/g)(0)=5/0" which is undefined"$

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Question #7ca7c

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