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liberstina [14]
1 year ago
12

4x2 + 4x - 1, evaluate and fully simplify each of the - For the function f(0) following f(s + h) f(x + h) - f(5) h 10

Mathematics
1 answer:
Andrew [12]1 year ago
7 0

Given the function f(x);

f(x)=-4x^2+4x-1

Evaluating the function f(x+h);

\begin{gathered} f(x+h)=-4(x+h)^2+4(x+h)-1 \\ f(x+h)=-4(x^2+2xh+h^2)^{}+4(x+h)-1 \\ f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1 \end{gathered}

So;

f(x+h)=-4x^2-4h^2-8xh^{}+4x+4h-1

Evaluating the second function;

\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1-(-4x^2+4x-1)}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2-4h^2-8xh^{}+4x+4h-1+4x^2-4x+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4x^2+4x^2-4h^2-8xh^{}+4x-4x+4h-1+1}{h} \\ \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h} \end{gathered}

simplifying further;

\begin{gathered} \frac{f(x+h)-f(x)}{h}=\frac{-4h^2-8xh^{}+4h}{h}=-4h-8x+4 \\ \frac{f(x+h)-f(x)}{h}=-4h-8x+4 \end{gathered}

Therefore, we have;

undefined

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4 step: Insert q= \frac{3}{2}into equation T_{1}*q^{3}*(q+1)=5 and obtain T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5, from where T_{1}= \frac{16}{27}.




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