Determine $f' (x)$ from first principles if $f(x) = -2x^2 - 1$ - NSC Mathematics - Question 7 - 2023 - Paper 1
Question 7
Determine $f' (x)$ from first principles if $f(x) = -2x^2 - 1$.
7.2 Determine:
7.2.1 $f' (x)$, if it is given that $f(x) = -2x^2 + 3x^2$
7.2.2 $\frac{dy}{dx}$ ... show full transcript
Worked Solution & Example Answer:Determine $f' (x)$ from first principles if $f(x) = -2x^2 - 1$ - NSC Mathematics - Question 7 - 2023 - Paper 1
Step 1
Determine $f' (x)$ from first principles if $f(x) = -2x^2 - 1$
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Answer
To find the derivative from first principles, we use the definition:
Substitute into the limit:
f′(x)=limh→0h−4xh−2h2=limh→0(−4x−2h)=−4x
Thus, f′(x)=−4x.
Step 2
Determine $f' (x)$, if it is given that $f(x) = -2x^2 + 3x^2$
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Answer
Firstly, simplify:
f(x)=(3−2)x2=x2
Now differentiate:
f′(x)=2x
Step 3
$\frac{dy}{dx}$ if $y = 2x + \frac{1}{\sqrt{4x}}$
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Answer
To find dxdy, we need to differentiate each term separately:
The derivative of 2x is 2.
For the second term:
y=2x+(4x)−1/2dxdy=2−21(4x)−3/2⋅4=2−(4x)3/22=2−2x3/21
So, dxdy=2−2x3/21.
Step 4
The graph $y = f' (x)$ has a minimum turning point at $(1; -3)$.
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Answer
To find where the function f is concave down, we need to determine the second derivative:
Find f′′(x):
Since f′(x)=−4x, differentiate again:
f′′(x)=−4
This indicates that f is concave down everywhere, as f′′(x)<0 for all x. Hence, f is concave down for all values of x.
