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h(x)={[(1)/(x-5),",",x=0],[-(x-4)^(2),",",x=7],[3x-2,",",x!=0]:}

h(x)={1x−5amp;,x=0−(x−4)2amp;,x=73x−2amp;,x≠0h(6)= \begin{array}{l}h(x)=\left\{\begin{array}{ll}\frac{1}{x-5} & , \quad x=0 \\ -(x-4)^{2} & , \quad x=7 \\ 3 x-2 & , \quad x \neq 0\end{array}\right. \\ h(6)= \\\end{array}

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Q. h(x)={1x−5,x=0−(x−4)2,x=73x−2,x≠0h(6)= \begin{array}{l}h(x)=\left\{\begin{array}{ll}\frac{1}{x-5} & , \quad x=0 \\ -(x-4)^{2} & , \quad x=7 \\ 3 x-2 & , \quad x \neq 0\end{array}\right. \\ h(6)= \\\end{array}
  1. Evaluate h(x)h(x) at x=0x = 0: The function h(x)h(x) is defined piecewise, meaning it has different expressions for different values of xx. We need to evaluate the function at x=0x = 0, x=7x = 7, and for x≠0x \neq 0.
  2. Evaluate h(x)h(x) at x=7x = 7: First, let's evaluate h(x)h(x) at x=0x = 0. According to the function definition, h(x)=1(x−5)h(x) = \frac{1}{(x-5)} when x=0x = 0. Let's plug in x=0x = 0 into this expression.\newlineh(0)=1(0−5)=1(−5)=−15h(0) = \frac{1}{(0-5)} = \frac{1}{(-5)} = -\frac{1}{5}.
  3. Evaluate h(x)h(x) for x≠0x \neq 0: Next, we evaluate h(x)h(x) at x=7x = 7. The function definition tells us that h(x)=−(x−4)2h(x) = -(x-4)^2 when x=7x = 7. Let's plug in x=7x = 7 into this expression.\newlineh(7)=−(7−4)2=−(3)2=−9h(7) = -(7-4)^2 = -(3)^2 = -9.
  4. Evaluate h(x)h(x) for x≠0x \neq 0: Next, we evaluate h(x)h(x) at x=7x = 7. The function definition tells us that h(x)=−(x−4)2h(x) = -(x-4)^2 when x=7x = 7. Let's plug in x=7x = 7 into this expression.\newlineh(7)=−(7−4)2=−(3)2=−9h(7) = -(7-4)^2 = -(3)^2 = -9.Lastly, we need to evaluate h(x)h(x) for x≠0x \neq 0. The function definition for this case is x≠0x \neq 000. This expression is valid for all x≠0x \neq 011 except x≠0x \neq 022. There is no specific value to calculate here, as the expression x≠0x \neq 033 is the value of h(x)h(x) for all x≠0x \neq 0.

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