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max of x^2+2y for b>4?

Try b=7 and see whether it is correct.

The 2 integrals at the ends may have different constants, so the conclusion is wrong

Try b=7

Sorry, no

Thanks. The relationship does change for x>=0. For x>ln2, d(e^x)/dx > d(2x)/dx. So I looked at each interval.

The relationship does change. At x=0, d(e^x)/dx < d(2x)/dx. At x=ln2, they are the same. So I pointed out x=0, e^0>2(0) AND x=ln2, 2>2ln2.

I thought I did that by pointing out at x=0, e^0>2(0).

How do you show that?

Why is it incomplete?

Thanks for that. In that case, just drop 2^x in the inequality. To justify e^x >2x: x=0, 1>0. d/dx of e^x = d/dx of 2x when x=ln2. When x=ln2, 2>2ln2. Since both are increasing, .: e^x>2x in [0,ln2] For x>ln2, d/dx of e^x > d/dx of 2x, .: e^x>2x.

The locus of point P(x,y) is (x/2)^2+(y/b)^2=1, b>0. Find, in terms of b, the maximum value of x^2+2y.

easy move the 3 from the denominator to replace the 1 at the numerator

Not worry about 'hence' e> 2\\e^{x}\geq 2^{x}\geq 2x\\.:\frac{d}{dx}e^{x}\geq \frac{d}{dx}\left ( 1+x^{2} \right ) At x=0,\\e^{x}=1+x^{2}=1,\\.:e^{x}\geq 1+x^{2} for x\geq 0

\left ( 1-x \right )^{2}\geq 0\\.: 1+x^{2}\geq 2x

I'll do it anyway: |z_1|≤|z_1 | true Assume |z_1+z_2+⋯+z_k |≤|z_1 |+|z_2 |+⋯+|z_k | true .: |z_1+z_2+⋯+z_k+1 |≤|z_1+z_2+⋯+z_k |+|z_k+1|≤|z_1 |+|z_2 |+⋯+|z_k | +|z_k+1| true .: |z_1+z_2+⋯+z_n |≤|z_1 |+|z_2 |+⋯+|z_n | true

So you are now saying you were talking nonsense when you said 'f(x) isn't a function; it's a rule'.

Impossible. Just try x=-1.

Here you said f(x) is a rule. The other post you said f(x) is a value. What is it?