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# help function

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Let f(x)=3x^2 - 4x - x^2 + 7x + 8. Find the constant k such that f(x)=f(k-x) for all real numbers x.

Jun 9, 2022

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Hi!
So first, simplify $$f(x)$$ to get:

$$f(x)=2x^2+3x+8$$

Then, find $$f(k-x)$$ as follows:

$$f(k-x)=2(k-x)^2+3(k-x)+8$$

$$=2(k^2-2kx+x^2)+3k-3x+8$$

$$=2x^2-4kx+2k^2+3k-3x+8$$

$$=2x^2+(-4k-3)x+(3k+8+2k^2)$$

Thus, $$f(k-x)=2x^2+(-4k-3)x+(2k^2+3k+8)$$

Then, set:

$$f(x)=f(k-x)$$

$$\iff 2x^2+3x+8=2x^2+(-4k-3)x+(2k^2+3k+8)$$

$$\iff 3x+8=(-4k-3)x+(2k^2+3k+8)$$

Equate coefficients of $$x:$$

$$3=-4k-3 \implies k=-\frac{3}{2}$$

Therefore, $$k=-\dfrac{3}{2}$$

Jun 9, 2022