Given f(x) = x^4 + 6x^3 - 15x + 7, evaluate \displaystyle \lim_{h \to 0} \frac{f(1+h) - f(1-h)}{h}.
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To evaluate the limit \displaystyle \lim_{h \to 0} \frac{f(1+h) - f(1-h)}{h}, we need to substitute the function f(x) and then simplify the expression.
Let's start by substituting the given function f(x) = x^4 + 6x^3 - 15x + 7 into the expression:
\displaystyle \lim_{h \to 0} \frac{(1+h)^4 + 6(1+h)^3 - 15(1+h) + 7 - [(1-h)^4 + 6(1-h)^3 - 15(1-h) + 7]}{h}
Now, let's simplify this expression step by step.
First, expand (1+h)^4 and (1-h)^4:
(1+h)^4 = 1 + 4h + 6h^2 + 4h^3 + h^4
(1-h)^4 = 1 - 4h + 6h^2 - 4h^3 + h^4
Next, expand (1+h)^3 and (1-h)^3:
(1+h)^3 = 1 + 3h + 3h^2 + h^3
(1-h)^3 = 1 - 3h + 3h^2 - h^3
Now, substitute these values back into the expression:
\displaystyle \lim_{h \to 0} \frac{(1 + 4h + 6h^2 + 4h^3 + h^4) + 6(1 + 3h + 3h^2 + h^3) - 15(1 + h) + 7 - [(1 - 4h + 6h^2 - 4h^3 + h^4) + 6(1 - 3h + 3h^2 - h^3) - 15(1 - h) + 7]}{h}
Simplifying further:
\displaystyle \lim_{h \to 0} \frac{1 + 4h + 6h^2 + 4h^3 + h^4 + 6 + 18h + 18h^2 + 6h^3 - 15 - 15h + 7 - (1 - 4h + 6h^2 - 4h^3 + h^4 + 6 - 18h + 18h^2 - 6h^3 - 15 + 15h + 7)}{h}
Now, combine like terms within the numerator:
\displaystyle \lim_{h \to 0} \frac{4h + 6h^2 + 4h^3 + 18h - 15h - 4h^3 - 18h^2 + 15h}{h}
Simplifying further gives:
\displaystyle \lim_{h \to 0} \frac{27h}{h}
Finally, cancel out the h terms:
\displaystyle \lim_{h \to 0} 27
Therefore, the limit \displaystyle \lim_{h \to 0} \frac{f(1+h) - f(1-h)}{h} is equal to 27.
To evaluate the limit \(\displaystyle \lim_{h \to 0} \frac{f(1+h) - f(1-h)}{h}\) , we start by substituting the expressions \( f(1+h) \) and \( f(1-h) \) into the numerator of the fraction.
\( f(1+h) = (1+h)^4 + 6(1+h)^3 - 15(1+h) + 7 \)
\( f(1-h) = (1-h)^4 + 6(1-h)^3 - 15(1-h) + 7 \)
Simplifying the expressions inside the parentheses, we have:
\( f(1+h) = (1+h)^4 + 6(1+h)^3 - 15(1+h) + 7 \)
\( = (1+4h+6h^2+4h^3+h^4) + 6(1+3h+3h^2+h^3) - 15(1+h) + 7 \)
\( = 1+4h+6h^2+4h^3+h^4 + 6+18h+18h^2+6h^3 - 15 - 15h + 7 \)
\( = 20 + 16h + 24h^2 + 10h^3 + h^4 \)
Similarly, simplifying \( f(1-h) \):
\( f(1-h) = (1-h)^4 + 6(1-h)^3 - 15(1-h) + 7 \)
\( = (1-4h+6h^2-4h^3+h^4) + 6(1-3h+3h^2-h^3) - 15(1-h) + 7 \)
\( = 1-4h+6h^2-4h^3+h^4 + 6-18h+18h^2-6h^3 - 15 + 15h + 7 \)
\( = 9 - 16h + 24h^2 - 10h^3 + h^4 \)
Now, substituting these expressions back into the original limit expression:
\( \displaystyle \lim_{h \to 0} \frac{f(1+h) - f(1-h)}{h} = \lim_{h \to 0} \frac{(20 + 16h + 24h^2 + 10h^3 + h^4) - (9 - 16h + 24h^2 - 10h^3 + h^4)}{h} \)
Simplifying the numerator:
\( = \lim_{h \to 0} \frac{20 + 16h + 24h^2 + 10h^3 + h^4 - 9 + 16h - 24h^2 + 10h^3 - h^4}{h} \)
\( = \lim_{h \to 0} \frac{11 + 32h + 20h^3}{h} \)
Now, dividing each term in the numerator by \( h \):
\( = \lim_{h \to 0} \frac{11}{h} + \lim_{h \to 0} \frac{32h}{h} + \lim_{h \to 0} \frac{20h^3}{h} \)
Simplifying each limit term:
\( = \lim_{h \to 0} 11 + \lim_{h \to 0} 32 + \lim_{h \to 0} 20h^2 \)
Finally, evaluating each term:
\( = 11 + 32 + 0 = 43 \)
Therefore, the limit \(\displaystyle \lim_{h \to 0} \frac{f(1+h) - f(1-h)}{h} = 43 \).